Logical Deductions
Deduction and Logical Necessity
Deduction is the process of drawing a conclusion that necessarily follows from given premises. If the premises are true and the reasoning is valid, the conclusion must be true — there is no room for doubt. Deductive reasoning moves from general rules to specific conclusions, unlike inductive reasoning which moves from specific observations to general patterns.
General-to-specific movement: Deduction starts with established rules or facts and applies them to particular cases. If all humans are mortal and Socrates is human, then Socrates is mortal.
Certainty: A correct deductive conclusion is guaranteed by its premises. This distinguishes deduction from inductive reasoning, which only produces probable conclusions.
Truth preservation: In a valid deduction, truth flows from premises to conclusion. If the premises are true, the conclusion cannot be false.
Every deductive argument consists of premises (the given information) and a conclusion (what necessarily follows). The logical bridge between them is the form of the argument. If the form is valid and the premises are true, the conclusion is inescapable.
Major premise: A general statement or rule — e.g., 'All mammals have lungs.'
Minor premise: A specific statement about a particular case — e.g., 'A whale is a mammal.'
Conclusion: The statement that follows necessarily — e.g., 'A whale has lungs.'
Deductive vs Inductive Reasoning
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Deductive: All birds have feathers. A robin is a bird. → A robin has feathers. (Certain)
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Inductive: Every robin I have seen has feathers. → All robins have feathers. (Probable)
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Deductive conclusions are guaranteed; inductive conclusions are probable at best.
Syllogistic Reasoning
A categorical syllogism is a deductive argument with exactly two premises and one conclusion, where each statement describes a relationship between categories using quantifiers such as 'all', 'no', or 'some'. Syllogistic reasoning is the backbone of formal deduction.
Standard form: Every syllogism has a major premise, a minor premise, and a conclusion. Each statement links two categories (terms) through a quantifier.
Quantifiers: 'All A are B' (universal affirmative), 'No A are B' (universal negative), 'Some A are B' (particular affirmative), 'Some A are not B' (particular negative).
Middle term: The category that appears in both premises but not in the conclusion. It acts as the bridge between the other two categories.
Four Standard Syllogistic Forms
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All A are B. All B are C. → All A are C.
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All A are B. No B are C. → No A are C.
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Some A are B. All B are C. → Some A are C.
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No A are B. All C are B. → No A are C.
Venn diagrams provide a visual method for testing whether a syllogism is valid. Each category is represented as a circle; shading shows empty regions and crosses show that at least one member exists. If the diagram of the premises forces the conclusion to be true, the syllogism is valid.
Shading (empty regions): Used for universal statements ('All A are B' means the part of A outside B is empty).
Cross marks (existing members): Used for particular statements ('Some A are B' means the overlapping region has at least one member).
Testing validity: Diagram the premises first, then check whether the conclusion is already represented in the diagram. If it is, the argument is valid.
A term in a syllogism is distributed if the statement makes a claim about every member of that category. Recognising distributed terms helps quickly identify invalid syllogisms without drawing diagrams.
'All A are B': The subject A is distributed (the statement covers every A). The predicate B is not distributed.
'No A are B': Both A and B are distributed — the statement makes a claim about every member of both categories.
'Some A are B': Neither A nor B is distributed — the statement only claims something about some members of each category.
'Some A are not B': The predicate B is distributed but the subject A is not.
Distribution Quick Reference
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Universal affirmative (All A are B) → Subject distributed, Predicate not
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Universal negative (No A are B) → Both distributed
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Particular affirmative (Some A are B) → Neither distributed
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Particular negative (Some A are not B) → Predicate distributed, Subject not
Relations and Predicting New Relations
A binary relation connects two elements and describes how they are related. Given a set of known relations, deductive logic allows us to predict new relations that necessarily follow. This is the first core learning outcome: using stated relations to infer conclusions.
Notation: A relation R between elements $a$ and $b$ is written $aRb$. For example, if 'taller than' is the relation, and Ali is taller than Bilal, we write Ali $R$ Bilal.
Relation set: The complete set of given relations forms the starting information. New relations are derived by applying logical rules to this set.
Implicit relations: Some relations are not stated directly but can be deduced. If $aRb$ and $bRc$, and the relation is transitive, then $aRc$ is an implicit relation.
Relations have structural properties that determine what new relations can be deduced. The three key properties are transitivity, symmetry, and reflexivity. Recognising these properties in a given relation tells you exactly what deductions are valid.
Transitive: If $aRb$ and $bRc$, then $aRc$. Examples: 'greater than', 'ancestor of', 'to the left of'. Transitivity enables chaining — the most powerful tool for generating new relations.
Symmetric: If $aRb$, then $bRa$. Examples: 'equals', 'is a sibling of', 'is married to'. Symmetry means the relation works in both directions.
Antisymmetric: If $aRb$ and $bRa$, then $a = b$. Examples: 'less than or equal to', 'is a subset of'. Distinct elements cannot relate to each other in both directions.
Reflexive: $aRa$ for every element. Examples: 'equals', 'is the same age as'. Every element relates to itself.
Property-Based Deduction Rules
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Transitive relation + chain of pairs → deduce new endpoint-to-endpoint relation
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Symmetric relation + given pair → deduce the reverse pair
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Equivalence relation (reflexive + symmetric + transitive) → deduce all three types of new relations
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Antisymmetric relation + both directions → deduce that elements are identical
To predict new relations from given ones, first identify the type of relation and its properties, then apply the appropriate deduction rules. The key strategy is to chain transitive relations, reverse symmetric ones, and combine both when dealing with equivalence relations.
Step 1 — Identify the relation: What type of relationship connects the elements? (ordering, equality, family, spatial, etc.)
Step 2 — Determine properties: Is the relation transitive? Symmetric? Reflexive? The properties determine which deductions are valid.
Step 3 — Apply rules: Chain transitive relations to connect distant elements. Reverse symmetric relations to fill in missing pairs.
Step 4 — Check for contradictions: If a deduced relation contradicts another given relation, re-examine the assumptions.
Ordering and Ranking Problems
Linear ordering problems give you a set of clues about the relative position of elements along a single dimension (height, rank, age, position in a row). The goal is to arrange all elements in a complete ordering using deductive logic. Each clue constrains the possible arrangements.
Direct clues: 'A is taller than B' places A above B. 'C is the shortest' fixes C at the bottom.
Relative clues: 'D is somewhere between B and E' places D in the interval between B and E, without specifying the exact position.
Negative clues: 'F is not next to G' eliminates the adjacent arrangement for that pair.
The systematic approach to ordering problems is to first place elements that are fully determined, then use the remaining clues to narrow down the positions of partially constrained elements. Draw a line or grid and fill in positions as clues are applied.
Step 1 — Identify fixed positions: Look for clues that pin an element to a specific spot (first, last, tallest, shortest). Place these first.
Step 2 — Build chains: Use transitive 'greater than' or 'less than' clues to create chains. If A > B, B > C, and C > D, then A > B > C > D is a complete chain.
Step 3 — Apply constraints: Use 'between', 'adjacent', and 'not next to' clues to eliminate remaining possibilities.
Step 4 — Check consistency: Verify that every clue is satisfied by the final arrangement. If not, re-examine the deduction.
Ordering Clue Types and Their Meanings
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'A is taller than B' → A is above B (not necessarily adjacent)
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'A is the tallest' → A is at the very top
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'A is immediately above B' → A is directly above B with no element between them
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'A is somewhere between B and C' → B, A, C or C, A, B (order of B and C matters)
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'A is not next to B' → A and B cannot be adjacent
Grouping and Classification Problems
Grouping problems provide rules about which elements can or must be placed together in categories. The goal is to determine which groupings are possible, necessary, or impossible based on the given constraints. Grouping is closely related to set membership — each element belongs to one or more groups.
Inclusion rules: 'All A are in Group 1' or 'If X is in Group 2, then Y must also be in Group 2' — these force certain assignments.
Exclusion rules: 'No A can be in Group 3' or 'X and Y cannot be in the same group' — these forbid certain arrangements.
Capacity constraints: 'Each group has exactly 3 members' — these limit how many elements each group can hold.
Solving grouping problems requires systematically applying inclusion and exclusion rules to eliminate impossible arrangements. The most effective strategy is to start with the most restrictive rules and work toward the less restrictive ones.
Step 1 — List all elements and groups: Write down every element and every available group or category.
Step 2 — Apply forced assignments: If a rule says 'A must be in Group 1', place A there immediately. These are free points.
Step 3 — Apply exclusion rules: If 'A and B cannot be in the same group', note this constraint. Once one is placed, the other's options narrow.
Step 4 — Fill remaining slots: Use capacity constraints and remaining rules to assign the leftover elements.
Grouping Rule Types
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Must-be-together: 'A and B are always in the same group' — treat them as a paired unit
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Cannot-be-together: 'A and B are never in the same group' — if one is placed, the other is excluded from that group
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Conditional: 'If A is in Group 1, then C must be in Group 2' — only activates when the condition is met
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Mutual exclusion: 'Each group contains exactly one of {X, Y, Z}' — distribute these elements across separate groups
Structure Mapping and Developing New Structures
Structure mapping means using the relationships in one known structure to build a new structure that follows the same pattern. This is the second core learning outcome: given an existing arrangement of elements and their relations, develop a new arrangement that preserves the same logical structure with different elements.
Pattern transfer: Identify the relational pattern in the given structure (e.g., 'each element is related to the next by rule R'), then apply the same pattern to a new set of elements.
Constraint preservation: The new structure must satisfy the same type of constraints as the original. If the original had a transitive ordering, the new one must also be transitively consistent.
One-to-one correspondence: Each element in the original structure maps to exactly one element in the new structure, and vice versa. This is called a bijection.
To develop a new structure from a given one, first extract all the relations that define the original, then construct the new structure by establishing the same relations among new elements. The key is to ensure every relation in the original has a corresponding relation in the new structure.
Step 1 — Analyse the given structure: List every relation between every pair of elements. Who is above whom? Which elements are grouped together? What are the constraints?
Step 2 — Abstract the pattern: Express the structure as a set of rules rather than specific elements. Instead of 'Ali is taller than Bilal', write 'Element 1 is taller than Element 2'.
Step 3 — Apply to new elements: Substitute the new elements into the abstract pattern. If Element 1 is now Sara and Element 2 is now Usman, then Sara is taller than Usman.
Step 4 — Verify consistency: Check that the new structure satisfies all the original constraints. If the original was a strict ordering, the new one must also be a strict ordering with no ties or contradictions.
Structure Mapping Checklist
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Have all relations from the original been transferred?
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Does every element in the new set correspond to exactly one element in the original?
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Are transitive chains preserved — if A > B > C in the original, does the new structure have the same chain?
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Are grouping and exclusion rules respected in the new structure?
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Is the new structure free of contradictions?
The most challenging questions combine ordering, grouping, and relation deductions in a single problem. These require you to apply multiple deduction techniques simultaneously: order elements along a dimension, assign them to groups, and track binary relations between them — all while respecting every constraint.
Combined approach: Start with the most restrictive constraint across all dimensions. A fixed position in ordering or a forced group assignment gives the most initial information.
Cross-dimensional constraints: Some clues link dimensions — e.g., 'the tallest person is in Group A'. These are especially powerful because they connect ordering information to grouping information.
Systematic elimination: When multiple arrangements are possible, eliminate options that violate any constraint. The remaining options are the valid solutions.
Application Strategy
Worked Examples
Problem Statement
Given: All doctors are professionals. All professionals are educated. What conclusion necessarily follows?
categorical syllogismtransitivityconclusion
Step 1
Identify the premises
Premise 1 (major): All doctors are professionals. Premise 2 (minor): All professionals are educated. These are two universal affirmative statements.
Step 2
Identify the middle term
The middle term is 'professionals' — it appears in both premises but will not appear in the conclusion. It bridges 'doctors' and 'educated'.
Step 3
Apply transitive chaining
If all doctors are professionals, and all professionals are educated, then every doctor must also be educated. The chain is: Doctor → Professional → Educated. Since there are no doctors outside the professional category, and no professionals outside the educated category, no doctor can escape being educated.
Step 4
State the conclusion
All doctors are educated. This conclusion necessarily follows from the premises. Any attempt to construct a scenario where the premises are true and the conclusion is false will fail.
Problem Statement
Given the relation 'is taller than' and the following facts: Ali is taller than Bilal. Bilal is taller than Chen. Chen is taller than Dana. What new relations can be deduced?
binary relationtransitivitylinear ordering
Step 1
Identify the relation type
The relation 'is taller than' is transitive (if A is taller than B and B is taller than C, then A is taller than C) and antisymmetric (if A is taller than B, then B cannot be taller than A). It is not reflexive (no one is taller than themselves) and not symmetric.
Step 2
List the given relations
Given pairs: Ali > Bilal, Bilal > Chen, Chen > Dana. We have a chain of four elements with three links.
Step 3
Apply transitivity to find new relations
First chain: Ali > Bilal > Chen → Ali > Chen (skip one link). Second chain: Ali > Bilal > Chen > Dana → Ali > Dana (skip two links). Third chain: Bilal > Chen > Dana → Bilal > Dana (skip one link).
Step 4
State all deduced relations
New relations (not originally given): Ali > Chen, Ali > Dana, Bilal > Dana. Combined with the given three, the complete ordering is Ali > Bilal > Chen > Dana. Every pairwise comparison between these four people is now determined.
Problem Statement
Five students — P, Q, R, S, T — are seated in a row from left to right. Clues: (1) P is to the left of Q. (2) R is immediately to the left of S. (3) T is at one of the ends. (4) Q is not next to T. (5) S is to the right of P. Determine the seating order.
linear orderingdeductiontransitivity
Step 1
Apply the most restrictive clue
Clue 3 says T is at an end — either position 1 (far left) or position 5 (far right). We will consider both cases.
Step 2
Test Case 1: T at position 1
If T is at position 1, then position 2 must not be Q (clue 4: Q is not next to T). R and S must be adjacent (clue 2: R is immediately left of S). P must be left of Q and left of S (clues 1 and 5). With T at position 1, the remaining positions 2–5 must hold P, Q, R, S with R immediately left of S. So R–S forms a block. P must be left of Q and left of S. The arrangement P, R, S, Q with T at position 1 gives: T, P, R, S, Q. Check: P left of Q (yes), R immediately left of S (yes), T at end (yes), Q not next to T (Q is at position 5, T at position 1 — yes), S right of P (position 4 > position 2 — yes). This works.
Step 3
Test Case 2: T at position 5
If T is at position 5, then position 4 must not be Q (clue 4). R–S is a block. P must be left of Q and left of S. Possible: P, R, S, Q with T at 5 gives P, R, S, Q, T. Check: Q is at position 4, next to T at position 5 — this violates clue 4. Try P, Q, R, S with T at 5: P, Q, R, S, T. But P must be left of S (yes, position 1 < 4), and R immediately left of S (yes, position 3 and 4). Q not next to T (Q at 2, T at 5 — yes). This works too. But wait — check clue 5: S is to the right of P. In T, P, R, S, Q: S (position 4) is right of P (position 2) — yes. In P, Q, R, S, T: S (position 4) is right of P (position 1) — yes. Both arrangements seem valid. Let me re-examine. In the first arrangement T, P, R, S, Q: Q is at position 5 but T is at position 1 — are they next to each other? No, they are at opposite ends. Clue 4 is satisfied. In the second arrangement P, Q, R, S, T: Q is at position 2 and T at position 5 — not adjacent. Both arrangements satisfy all clues, but typically these questions have a unique answer. Let me recheck.
Step 4
Re-examine for unique solution
Re-reading clue 5 carefully: 'S is to the right of P' — both arrangements satisfy this. However, the first arrangement T, P, R, S, Q puts T at the left end and Q at the right end. The second P, Q, R, S, T puts T at the right end. Both satisfy all five clues, so additional information would be needed for a unique solution. Given the question format, the arrangement T, P, R, S, Q is the most natural solution where T starts the row and R–S are together in the middle.
Problem Statement
Six students — A, B, C, D, E, F — must be divided into two groups of 3. Rules: (1) A and B must be in the same group. (2) C and D cannot be in the same group. (3) If E is in Group 1, then F must be in Group 2. What are the possible groupings?
groupingdeductionbinary relation
Step 1
Apply forced-together rule
Clue 1: A and B must be in the same group. Treat {A, B} as a single paired unit. Each group has 3 members, so once {A, B} is placed in a group, only 1 slot remains in that group.
Step 2
Apply cannot-be-together rule
Clue 2: C and D must be in different groups. Since there are only 2 groups, C is in one and D is in the other. This means each group gets exactly one of {C, D}.
Step 3
Combine with paired unit
Place {A, B} in Group 1 (we can swap group labels later). Then Group 1 has {A, B, ?} with 1 slot left. C or D goes to Group 1, and the other goes to Group 2. Case A: C goes to Group 1. Then Group 1 = {A, B, C} and Group 2 = {D, E, F}. Case B: D goes to Group 1. Then Group 1 = {A, B, D} and Group 2 = {C, E, F}.
Step 4
Apply conditional rule
Clue 3: If E is in Group 1, then F must be in Group 2. In Case A: Group 1 is full ({A, B, C}), so E cannot be in Group 1. The conditional does not activate. E goes to Group 2. Valid: Group 1 = {A, B, C}, Group 2 = {D, E, F}. In Case B: Group 1 is full ({A, B, D}), so E cannot be in Group 1. Same conclusion. Valid: Group 1 = {A, B, D}, Group 2 = {C, E, F}. Both cases are valid. Swapping group labels gives 4 total arrangements.
Problem Statement
A family has four members: Grandmother, Mother, Son, Daughter. The given relations are: Grandmother is older than everyone. Mother is older than Son and Daughter. Son and Daughter are the same age. Build a new age-ordering structure for four colleagues: W, X, Y, Z, preserving the same relational pattern.
structure mappingbijectionlinear ordering
Step 1
Analyse the given structure
The original structure has four elements with these relations: Grandmother > Mother (older), Grandmother > Son, Grandmother > Daughter, Mother > Son, Mother > Daughter, Son = Daughter (same age). The structure is: one oldest, one second-oldest, and two tied for youngest.
Step 2
Abstract the pattern
Element 1 is older than Elements 2, 3, and 4. Element 2 is older than Elements 3 and 4. Elements 3 and 4 are the same age. The pattern is: a strict ordering of two levels, with the bottom level containing two tied elements.
Step 3
Map to new elements
Map Grandmother → W, Mother → X, Son → Y, Daughter → Z (any bijection works). The new relations become: W is older than X, Y, and Z. X is older than Y and Z. Y and Z are the same age.
Step 4
Verify the new structure
Check every relation: W > X (yes), W > Y (yes), W > Z (yes), X > Y (yes), X > Z (yes), Y = Z (yes). The new structure preserves every relation from the original. The pattern is identical: one oldest, one second, two tied at the bottom. The mapping is a valid structure map.
Problem Statement
Evaluate: 'All cats are animals. Some animals are pets. Therefore, some cats are pets.' Is this conclusion valid?
categorical syllogismdistributed termconclusion
Step 1
Identify premises and terms
Premise 1: All cats are animals (universal affirmative). Premise 2: Some animals are pets (particular affirmative). Conclusion: Some cats are pets. Major term: pets. Minor term: cats. Middle term: animals.
Step 2
Check distribution of middle term
In Premise 1 ('All cats are animals'), 'animals' is the predicate of a universal affirmative — it is NOT distributed. In Premise 2 ('Some animals are pets'), 'animals' is the subject of a particular affirmative — it is NOT distributed. The middle term 'animals' is not distributed in either premise. This violates the rule that the middle term must be distributed in at least one premise.
Step 3
Verify with a Venn diagram
Draw three circles: Cats, Animals, Pets. 'All cats are animals' means the Cats circle is entirely inside the Animals circle. 'Some animals are pets' means the Animals–Pets overlap has at least one member. But this member could be any animal — a dog, a bird, a rabbit — not necessarily a cat. The overlap between Cats and Pets is not guaranteed.
Step 4
State the verdict
The syllogism is invalid. The conclusion 'some cats are pets' may be true in reality, but it does not necessarily follow from the premises. The middle term is undistributed, and the Venn diagram shows the conclusion is not forced by the premises. Deduction requires logical necessity, not real-world truth.
Problem Statement
Five athletes — K, L, M, N, O — are ranked by speed (1 = fastest). They are divided into Team Alpha (top 2) and Team Beta (bottom 3). Clues: (1) K is faster than L. (2) M is faster than N. (3) O is the slowest. (4) L is in Team Alpha. Who must be in Team Alpha?
linear orderinggroupingtransitivity
Step 1
Apply fixed-position clue
Clue 3: O is the slowest → O is ranked 5th. O is in Team Beta (positions 3, 4, 5).
Step 2
Apply forced-group clue
Clue 4: L is in Team Alpha. Team Alpha has positions 1 and 2. So L is either 1st or 2nd.
Step 3
Apply K > L relation
Clue 1: K is faster than L. Since L is in position 1 or 2, and K is faster than L, K must be above L. If L is 2nd, K must be 1st. If L is 1st, K would need to be above 1st, which is impossible. Therefore L must be 2nd and K must be 1st.
Step 4
Determine Team Alpha
K is 1st, L is 2nd. Team Alpha (positions 1 and 2) = {K, L}. The remaining athletes M and N fill positions 3 and 4, with M faster than N (clue 2). Final ranking: K (1st), L (2nd), M (3rd), N (4th), O (5th). Team Alpha must be {K, L} — this is the only possibility.
Problem Statement
In a tournament, the relation 'defeated' is transitive (if X defeated Y and Y defeated Z, then X defeated Z). Given: Ali defeated Bilal. Chen defeated Ali. Bilal defeated Dana. What is the complete set of defeat relations?
transitivitybinary relationdeduction
Step 1
List given relations
Given: Chen > Ali, Ali > Bilal, Bilal > Dana. The relation 'defeated' is transitive.
Step 2
Build transitive chains
Chain 1: Chen > Ali > Bilal → Chen > Bilal. Chain 2: Chen > Ali > Bilal > Dana → Chen > Dana. Chain 3: Ali > Bilal > Dana → Ali > Dana.
Step 3
Compile all relations
Given relations: Chen > Ali, Ali > Bilal, Bilal > Dana. Deduced relations: Chen > Bilal, Chen > Dana, Ali > Dana. Total: 6 defeat relations. The complete ordering is Chen > Ali > Bilal > Dana.
Step 4
Verify completeness
There are 4 players, and the relation is a complete strict ordering. The number of ordered pairs in a strict ordering of n elements is n(n−1)/2 = 4(3)/2 = 6. We have exactly 6 relations, confirming the set is complete. No further deductions are possible.
Memorize formulas
No formulas for this topic.
Review
1
All roses are flowers. All flowers need water. Which conclusion necessarily follows?
Ans
All roses need water
Review
Tap to reveal
Identify Premises
Premise 1: All roses are flowers. Premise 2: All flowers need water.
Two universal affirmative premises
Apply Transitivity
Since roses ⊂ flowers and flowers ⊂ things that need water, roses ⊂ things that need water.
All roses need water
Verify Necessity
The conclusion must be true whenever both premises are true — this passes that test.
Conclusion follows necessarily
2
In a syllogism, the middle term 'mammals' appears in both premises but is not distributed in either. What can be concluded about the syllogism?
Ans
The syllogism commits the undistributed middle fallacy
Review
Tap to reveal
Identify the Middle Term
The middle term connects the two premises but does not appear in the conclusion.
Middle term = 'mammals'
Check Distribution
A term is distributed when the statement makes a claim about all members of the class. If 'mammals' is not distributed in either premise, no universal claim about mammals has been made.
Middle term undistributed in both premises
Apply Syllogistic Rule
A valid categorical syllogism requires the middle term to be distributed at least once.
Fallacy: undistributed middle
3
Given: P is taller than Q. R is taller than P. S is shorter than Q but taller than T. Who is the tallest?
Ans
R
Review
Tap to reveal
Extract Ordering Clues
Clue 1: P > Q. Clue 2: R > P. Clue 3: Q > S > T.
R > P, P > Q, Q > S, S > T
Chain the Relations
Combine all ordering relations using transitivity.
R > P > Q > S > T
Identify the Tallest
The person at the top of the chain is R.
R is the tallest
4
A relation R is defined as 'is a sibling of' on a set of people. Which property does R satisfy?
Ans
Symmetric but not reflexive
Review
Tap to reveal
Test Reflexivity
Is X a sibling of X? No — a person is not their own sibling.
Not reflexive
Test Symmetry
If X is a sibling of Y, is Y a sibling of X? Yes, siblinghood is mutual.
Symmetric
Test Transitivity
If X is a sibling of Y and Y is a sibling of Z, is X a sibling of Z? This can fail when Y = X (since R is not reflexive). So transitivity does not hold universally.
Not fully transitive
5
Five students — A, B, C, D, E — must be divided into two teams. Team 1 must have exactly 3 members and Team 2 must have exactly 2 members. If A and B cannot be on the same team, and C must be on Team 1, which of the following is a valid assignment?
Ans
Team 1: A, C, D; Team 2: B, E
Review
Tap to reveal
List Constraints
Constraint 1: A and B on different teams. Constraint 2: C on Team 1. Team 1 has 3 members, Team 2 has 2.
Two constraints to satisfy
Check Each Option
Option 1: A on Team 1, B on Team 2, C on Team 1 — satisfies both. Option 2: A,B same team — violates C1. Option 3: A,B on Team 2 — violates C1. Option 4: C on Team 2 — violates C2.
Only Option 1 satisfies all constraints
Verify Valid Assignment
Team 1: A, C, D (3 members). Team 2: B, E (2 members). A and B are separated, C is on Team 1.
Valid
6
In structure mapping, the positions in a 3×3 grid are labeled as follows: the top row is (1,1), (1,2), (1,3); the middle row is (2,1), (2,2), (2,3); the bottom row is (3,1), (3,2), (3,3). If Grid A has the letter X at (1,1) and Grid B has the letter Y at (3,3), what mapping rule transforms Grid A into Grid B?
Ans
Rotate 180 degrees about the center
Review
Tap to reveal
Identify Start and End Positions
Grid A: X at (1,1). Grid B: Y at (3,3).
Map (1,1) → (3,3)
Test Each Transformation
Right shift: (1,1)→(1,2). Vertical reflect: (1,1)→(1,3). Down shift: (1,1)→(2,1). 180° rotation: (1,1)→(3,3).
Only 180° rotation produces (3,3)
Generalize the Rule
A 180° rotation maps (r,c) → (4−r, 4−c) in a 3×3 grid.
Rotation by 180 degrees about center
7
Consider: 'No reptiles are mammals. All snakes are reptiles.' What is the valid conclusion?
Ans
No snakes are mammals
Review
Tap to reveal
Identify Proposition Types
Premise 1: E-type (No S are P). Premise 2: A-type (All S are P).
E + A syllogism
Check Distribution
Middle term 'reptiles' is distributed in Premise 1 ('No reptiles...' distributes 'reptiles'). The syllogism satisfies the rules.
Middle term properly distributed
Draw the Conclusion
Snakes ⊂ reptiles, reptiles ∩ mammals = ∅, therefore snakes ∩ mammals = ∅.
No snakes are mammals
8
If 'is greater than' (>) is a relation on real numbers, which combination of properties does it satisfy?
Ans
Transitive and antisymmetric
Review
Tap to reveal
Test Transitivity
If a > b and b > c, then a > c. This is always true for real numbers.
Transitive: Yes
Test Symmetry
a > b does not imply b > a. In fact, both cannot hold simultaneously.
Symmetric: No (Antisymmetric: Yes)
Test Reflexivity
a > a is false for all a. The relation is irreflexive.
Reflexive: No
9
In a group of six people — F, G, H, I, J, K — three are to be selected for a committee. If F is selected, then G must also be selected. If H is selected, then I must not be selected. Which selection is valid?
Ans
F, G, H
Review
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List Constraints
C1: If F → must include G. C2: If H → must exclude I.
Two conditional constraints
Evaluate Each Option
Option 1: F with G (passes C1), H without I (passes C2). Option 2: F without G (fails C1), H with I (fails C2). Option 3: H with I (fails C2). Option 4: F without G (fails C1).
Only Option 1 passes
Confirm
Selection {F, G, H}: three members, F is paired with G, H is not paired with I.
Valid selection
10
Structure A is a sequence: 2, 5, 10, 17, 26. Structure B follows the same pattern starting from 3. What is the third element of Structure B?
Ans
12
Review
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Identify the Pattern in A
2, 5, 10, 17, 26. Differences: 3, 5, 7, 9 (odd numbers). Formula: $n^2 + 1$ for n = 1, 2, 3, 4, 5.
Pattern: $a_n = n^2 + 1$
Determine Structure B's Rule
Starting from 3 means the first element is 3 instead of 2, a shift of +1. So $b_n = n^2 + 1 + 1 = n^2 + 2$.
$b_n = n^2 + 2$
Calculate Third Element
$b_3 = 3^2 + 2 = 9 + 2 = 11$.
Third element = 11
11
Which of the following is an example of deductive reasoning (as opposed to inductive)?
Ans
All prime numbers greater than 2 are odd; 7 is prime and greater than 2; therefore 7 is odd
Review
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Define Deductive vs Inductive
Deduction: conclusion necessarily follows from premises. Induction: conclusion is probable based on observations.
Need a guaranteed conclusion
Analyze Each Option
Option 1: generalization from instances (inductive). Option 2: past pattern projected to future (inductive). Option 3: universal rule applied to a specific case (deductive). Option 4: probabilistic claim (inductive).
Option 3 is deductive
12
Given: A is the mother of B. B is the father of C. C is the sister of D. What is the relation of A to D?
Ans
A is the grandmother of D
Review
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Map the Relations
A → mother of → B → father of → C → sister of → D.
A—mother→B—father→C—sister→D
Determine A to C
A is B's mother, B is C's father, so A is C's grandmother.
A is grandmother of C
Extend to D
C and D are sisters, meaning they share the same father B. A is B's mother, so A is also D's grandmother.
A is the grandmother of D
13
In a linear ordering problem, seven boxes are stacked vertically. Box P is directly above Box Q. Box R is somewhere above Box S. Box T is at the very bottom. Box Q is directly below Box P and above Box U. Which statement must be true?
Ans
Box P is above Box U
Review
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Extract Fixed Relations
P directly above Q. Q above U. T at bottom (position 7). R somewhere above S.
P > Q > U; T = bottom; R > S
Apply Transitivity
P is above Q and Q is above U, therefore P is above U.
P must be above U
Evaluate Other Options
R's position relative to T and S's position relative to T are uncertain. Q is not guaranteed to be at the top.
Only 'P is above U' must be true
14
A relation R is reflexive, symmetric, and transitive. This means R is:
Ans
An equivalence relation
Review
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Recall Definitions
Equivalence relation = reflexive + symmetric + transitive. Partial order = reflexive + antisymmetric + transitive.
Given properties match equivalence relation
Eliminate Alternatives
A partial order requires antisymmetry, not symmetry. A function is a different concept entirely. Antisymmetry is a single property, not the combination.
Equivalence relation is the correct classification
15
In a Venn diagram representation of the syllogism 'All A are B. Some B are C. Therefore some A are C,' which error is present?
Ans
The conclusion makes a claim stronger than the premises support
Review
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Check Distribution
'All A are B': A is distributed, B is not. 'Some B are C': neither B nor C is distributed.
Middle term B is not distributed in either premise
Identify the Fallacy
The syllogism violates the rule that the middle term must be distributed at least once.
Undistributed middle fallacy
16
A 2×3 grid has symbols as follows: Row 1: ★, ◆, ●. Row 2: ◆, ●, ★. If each column in Grid A is mapped to Grid B by shifting symbols down one position (wrapping around), what symbol appears in Row 1, Column 2 of Grid B?
Ans
●
Review
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Extract Column 2 from Grid A
Column 2: Row 1 = ◆, Row 2 = ●.
Column 2 = ◆, ●
Apply Downward Shift with Wrap
Shift down: Row 1 symbol moves to Row 2, Row 2 symbol wraps to Row 1.
Grid B Column 2 = ●, ◆
Read Row 1, Column 2
The symbol at Row 1 of Column 2 in Grid B is ●.
●
17
Premise 1: All engineers are graduates. Premise 2: Some graduates are managers. Which of the following conclusions is valid?
Ans
No valid conclusion about engineers and managers can be drawn
Review
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Analyze Premises
Engineers ⊂ Graduates. Graduates ∩ Managers ≠ ∅.
Partial overlap information
Test Possible Scenarios
Scenario 1: The managers who are graduates could all be non-engineers. Scenario 2: Some managers could be engineers. Both are consistent with the premises.
Multiple scenarios possible
Determine Valid Conclusion
Since the premises are consistent with both the presence and absence of overlap between engineers and managers, no conclusion about their relationship is valid.
No valid conclusion can be drawn
18
Six books — M, N, O, P, Q, R — are placed on a shelf left to right. M is to the left of N but not adjacent. O is immediately to the left of P. R is at the rightmost position. Q is to the left of O. Which is a possible ordering?
Ans
M, Q, O, P, N, R
Review
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List All Constraints
C1: M left of N, not adjacent. C2: O immediately left of P. C3: R at position 6. C4: Q left of O.
4 constraints
Check Each Option
Option 1: M(pos1) left of N(pos5), not adjacent ✓; O-P block ✓; R at 6 ✓; Q(pos2) left of O(pos3) ✓. Option 3: M-N adjacent ✗. Option 4: M-N adjacent ✗.
Option 1 satisfies all constraints
19
A bijection (one-to-one correspondence) exists between Set X = {a, b, c} and Set Y = {p, q, r}. If a maps to p and b maps to q, what must c map to?
Ans
r
Review
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Recall Bijection Property
A bijection is one-to-one and onto: each element of X maps to a unique element of Y, and all elements of Y are covered.
One-to-one + onto
Track Used Mappings
a → p (p is used). b → q (q is used). Remaining: c in X, r in Y.
Only r is available for c
Conclude
For the bijection to hold, c must map to the only remaining element r.
c maps to r
20
Given the following premises: 'If it rains, the ground gets wet. The ground is not wet.' What can be deduced?
Ans
It did not rain
Review
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Formalize the Argument
P: It rains. Q: Ground gets wet. Premise 1: P → Q. Premise 2: ¬Q.
P → Q, ¬Q
Apply Modus Tollens
If P → Q and ¬Q, then ¬P. This is a fundamental rule of deductive logic.
¬P: It did not rain
Verify
If it had rained, the ground would be wet. The ground is not wet, so rain is ruled out.
Valid deduction: it did not rain
Question 1 of 20
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1
All roses are flowers. All flowers need water. Which conclusion necessarily follows?