Logical Deduction, Induction and Syllogisms

Syllogism with some and all

A syllogism is a deductive argument with two premises and one conclusion, in which the premises share a middle term that links the two other terms. In the some-and-all type, every statement takes one of four standard forms: All S are P (A), No S is P (E), Some S are P (I), and Some S are not P (O). In set notation, A means S \subseteq P, E means S \cap P = \varnothing, I means S \cap P \neq \varnothing, and O means S \not\subseteq P. Each conclusion must be checked against these forms before it is accepted.

The most important habit is to read 'some' in its logical sense. 'Some' means at least one, and it does not exclude 'all', so a statement with 'some' is still true when every member of the set has the property. By contrast, 'only some' and 'not all' assert that at least one member lacks the property, so a conclusion stating 'only some' is contradicted by any premise that says 'all'.

• at least one — one or more, possibly all. 예) At least one dancer in the troupe is a singer, and every dancer might be one.
• only some / not all — a strict partial claim that excludes 'all'. 예) Not all painters are sculptors, so only some painters are sculptors.
• almost all / most — allows exceptions, but if the group is nonempty it still yields 'some'. 예) Most of the members voted, so some members voted.
• either ... or — a disjunctive conclusion that is necessarily true when one of the two alternatives must hold.
• cannot be true simultaneously — the wording for a contrary or contradictory pair.
• logically inferred with certainty — must hold in every arrangement of the sets that satisfies the premises.

Conversion is the exchange of subject and predicate. The E and I forms convert simply: from No S is P we get No P is S, and from Some S are P we get Some P are S. For example, Some engineers are writers yields Some writers are engineers, and No writer is an actor yields No actor is a writer. The A form does not convert simply. From All S are P we may infer Some P are S only when S is known to be nonempty, and we can never infer All P are S, so any reversed universal conclusion is invalid.

When two premises share a middle term, the conclusion links the outer terms. All A are B and All B are C give All A are C. Some A are B together with All B are C gives Some A are C, because the A-members that are B are also C. The conclusion keeps the weaker quantifier 'some' and must not be strengthened to 'all'. Thus, if All X are Y and All Y are Z, then a conclusion that only some X are Z is contradicted by the premises.

When a particular premise is combined with a universal one, the chain works the same way. Some soaps are clean and All clean objects are wet give both Some clean objects are soaps and Some wet objects are soaps. Similarly, Some bottles are cups and All cups are knives give Some bottles are knives and Some knives are cups. A statement such as All creatures with claws are ferocious, together with the existence of clawed creatures, gives Some ferocious creatures have claws, but it never gives All creatures with claws are animals.

When a universal premise is paired with an E premise on another term, the result is an E conclusion. All teachers are professors and No professor is male give No teacher is male. A particular premise about males and engineers, however, says nothing about whether any engineer is a professor, so conclusions linking engineers and professors cannot be drawn.

The A and E forms are contraries: if the subject class is nonempty, All S are P and No S is P cannot both be true. The I and O forms are subcontraries and can both be true. The pairs A with O and E with I are contradictories, so exactly one of each pair holds. For instance, All students are inquisitive and No student is inquisitive cannot both be true when at least one student exists, while Some students are inquisitive and Some students are not inquisitive are compatible.

A conclusion of the form 'either I or II' is valid when the two conclusions are contradictory and exhaustive. Some black things are purple and No black thing is purple are I and E statements with the same subject and predicate, so one of them must hold in every arrangement. Premises that place black and purple inside a larger set decide neither conclusion alone, so each is not established, but their disjunction is necessarily true.

If a statement is given as false, its contradictory is true. The negation of Some S are not P is All S are P. Once this is established, Some S are P follows when S has at least one member, and any conclusion that denies the universal, such as No S is P, is contradicted.

Syllogism with some and all

A syllogism is a deductive argument in which a conclusion is drawn from two statements called premises. In GATE General Aptitude, the premises are usually built from the quantifiers all, some and no, and the question asks which conclusion follows with certainty. To answer correctly, you must read each quantifier in its exact logical sense.

There are four standard forms. Each has a fixed negation that you must know:

• All S are P (A): every member of S belongs to P. Its negation is 'Some S are not P.'
• No S are P (E): no member of S belongs to P. Its negation is 'Some S are P.'
• Some S are P (I): at least one member of S belongs to P. In this exam, some means 'at least one, possibly all', so it does not exclude 'all.' Its negation is 'No S are P.'
• Some S are not P (O): at least one member of S lies outside P. Its negation is 'All S are P.'

A particular affirmative can be converted simply. 'Some S are P' gives 'Some P are S', because the same individuals belong to both classes. The E statement also converts: 'No S are P' gives 'No P are S'. An A statement does not convert. 'All pilots are trained' does not give 'All trained people are pilots.'

A statement that says nothing about a class cannot be used to draw conclusions about that class. If the premise is 'Some humans are intelligent', nothing is said about all humans or about non-humans, so neither 'All humans are intelligent' nor 'Some non-humans are intelligent' can be inferred.

When a statement is declared false, you must assert its contradictory. If 'Some S are not P' is false, then 'All S are P' is true. The contradictory of 'All S are P' is 'Some S are not P,' which is the same as 'There exists at least one S that is not P.' For example, if 'All the boxes in the store are sealed' is false, then at least one box in the store is not sealed.

Universal statements are often read as presupposing that their subject class is not empty. Under the convention used in the keyed answers of this exam, 'All S are P' together with the existence of S yields 'Some S are P' and, by conversion, 'Some P are S.' This is why the falsity of 'Some S are not P' supports 'Some S are P' and 'Some P are S' when the subject exists. Formal logic does not make this assumption, so the answer depends on the exam convention.

If the falsity of 'Some S are not P' is given, then 'No S are P' does not follow, because it is the contrary of 'All S are P,' not its contradictory. Options that assert the opposite of a conclusion already established must be rejected.

A chain of universal statements gives a universal conclusion. 'All A are B' and 'All B are C' give 'All A are C.' The converse, 'All C are A,' is not implied. Reversing a universal statement is a common error. The same reasoning applies to the chain 'All heroes are winners' and 'All winners are lucky people,' which gives heroes inside lucky people but does not make every lucky person a hero.

Mixing a particular premise with a universal one gives a particular conclusion. If some creatures have claws and all creatures with claws are ferocious, then some creatures are ferocious and clawed. Converting the particular statement gives 'Some ferocious creatures have claws.' Statements about animals that were not mentioned, or about non-ferocious creatures, cannot be added.

The word almost is a qualifier that weakens 'all.' 'Almost all S are P' allows exceptions, so 'All S are P' cannot be drawn. It still presupposes a non-empty group, so 'Some S are P' follows. Statements about people outside the group cannot be drawn.

In eligibility questions, a condition such as 'Only those who completed a process are eligible' means that anyone who did not complete it is not eligible. If no member of a group completed the process, then no member of that group is eligible, so every eligible person lies outside the group. Ineligibility can have other causes, so 'All ineligible people belong to that group' does not follow.

A conditional statement 'If X, then Y' is logically equivalent to its contrapositive, 'If not Y, then not X,' and the contrapositive is necessarily true. The converse, 'If Y, then X,' and the inverse, 'If not X, then not Y,' are not implied. The phrase 'Y only when X' reverses the direction and means 'If Y, then X.' For example, 'When it rains, the match is cancelled' gives 'If the match is not cancelled, it is not raining,' but not 'If the match is cancelled, it is raining.'

Sets of categories are shown by circles. Three basic relations are disjoint (no common member), overlapping (some common members, neither contains the other) and nested (one is entirely inside the other). Whether named types are disjoint, overlapping or nested depends on whether they can share members. For example, cats and dogs as domestic animals are disjoint; a teacher can also be a poet, so the two overlap; every rose is a flower and every flower is a plant, so the sets are nested.

Counting questions apply the same logic to sets. If every member of C also belongs to B, then C is a subset of B, so its members are already inside the count of B and must not be added again.

If C and D have no common member, then |C \cap D| = 0. In that case the members of C can be removed from the total of D without double-counting.

A set split into mutually exclusive parts has a size equal to the sum of the parts: |B| = |B \cap C| + |B \setminus C| when only these two parts exist, and an unknown part equals the total minus the known parts.

A practical method is to subtract step by step. First remove the known subset, then remove the members with no further attribute, and the remainder is the overlap sought. For a union of two sets, inclusion–exclusion gives |A \cup B| = |A| + |B| - |A \cap B|, which yields the same overlap when the union is given.

In every case, verify each candidate conclusion against the stated premises. A conclusion that cannot be traced to a premise is not certain, however plausible it looks.

Figure-transformation questions ask you to identify one rule from a worked example and then apply it to new input. For instance, a reflection in a horizontal axis flips the rows but keeps the left-to-right order of the entries. The rule must hold for every entry, not only the first.

Syllogism with some and all

In logic, 'All A are B' means that every member of A is also a member of B, while 'No A are B' means that A and B have no common member, so the two groups are disjoint. 'Some A are B' means that at least one member of A belongs to B. It does not exclude the possibility that all members of A belong to B. A conclusion that depends on 'some' is therefore valid only if it holds in every arrangement consistent with the premises.

Counting problems of this kind are solved by partitioning a group into disjoint parts. When one group lies entirely inside another, the difference in their sizes gives the part of the larger group outside the smaller one, and a group that is disjoint from the overlap adds nothing to it. For example, if every member of X lies in Y, with |Y| = 50 and |X| = 20, then 50 - 20 = 30 members of Y lie outside X. For two overlapping conditions, the general identity is |A \cup B| = |A| + |B| - |A \cap B|, which is used when the number of people with at least one condition is given.

Note that 'Some A are B' does not imply 'Some A are not B'. 'Not all A are B' is equivalent to 'Some A are not B', and this equivalence is a frequent source of error. Before accepting a computed value, check that it is consistent with every premise.

Art gallery guard visibility

Puzzles of this type are graph problems in disguise. Each junction is a vertex and each walkway is an edge joining two junctions. A guard placed at a vertex watches every edge incident to that vertex, so the goal is to choose the fewest vertices such that every edge has at least one chosen endpoint. This is the minimum vertex cover problem, and the answer is the size of that cover.

A simple lower bound comes from vertex degrees. If the largest number of walkways meeting at one junction is \Delta and there are m walkways in total, then at least \lceil m/\Delta \rceil guards are needed, because each guard covers at most \Delta walkways. In a closed loop of five walkways every junction has degree 2, so the bound is \lceil 5/2 \rceil = 3. Two guards can cover at most four walkways in such a loop, so they are never enough.

For a cycle C_n the vertex cover number is \lceil n/2 \rceil. A loop of five junctions therefore needs three guards, and a loop of four junctions needs two. A lower bound alone does not prove that a set of that size exists, so the answer is confirmed by exhibiting a cover of that size. For an odd loop, choosing every other junction and one extra junction achieves the bound.

Distinguish a minimal cover from a minimum cover. A cover is minimal if no guard can be removed without leaving a walkway unwatched, but a minimal cover can still be larger than the minimum. Since the question asks for the minimum, a set of four guards that already contains a cover of three is not the answer.

Age ordering from statements

Comparative statements such as 'older than' and 'younger than' are transitive: if A is older than B and B is older than C, then A is older than C. They are also asymmetric, so if A is older than B, then B cannot be older than A. Convert each statement into an inequality with the correct direction before combining them. 'P is older than R' means that R is younger than P, and reversing this direction is a common trap.

Once every statement has been converted, combine the inequalities into a single chain running from youngest to oldest. A statement about one speaker's position, such as 'I am neither the youngest nor the oldest', does not fix a position directly. It excludes that person from both extremes, which means that at least one person is younger and at least one is older.

For example, if Anil is older than Bina and Chirag is younger than Bina, the chain is Chirag < Bina < Anil, so Chirag is the youngest. The method is to find the person who is forced to be below every other person in every consistent ordering. A candidate is excluded from the youngest position as soon as a statement places someone younger than him or her.

A conclusion is valid only if it is forced. Check each candidate ordering against all statements, and accept an answer only if it holds in every ordering that satisfies all the statements.

Compass direction reasoning

Direction problems rely on fixed reference points. The sun rises in the east and sets in the west, so in the morning a shadow points roughly towards the west. A shadow falling on a person's right side therefore indicates that the person's right-hand direction is west. A person facing south has their right hand to the west, a person facing north has their right hand to the east, a person facing east has their right hand to the south, and a person facing west has their right hand to the north.

When a person comes out of a building through its front door, the building lies behind them and the person faces the direction the front of the building faces. Reasoning in steps works well here: identify the shadow direction from the sun's position, convert it into a side of the person, deduce the person's facing direction, and then read off the building's orientation.

On a map drawn with north at the top, a point is northwest of a reference point only if it is both north and west of it. 'To the southeast of' means south and east together, not merely east. Two places that are separated by a road crossing lie in different quadrants of the crossing, and the quadrant determines the compass relation. Two places in the same row are simply east or west of each other. 'Next to' means adjacent with no separating road between them.

In ordering problems, a position is determined by counting. If k items lie east of a given item, that item occupies the (k+1)-th position from the east. 'To the west of' means anywhere further west, not necessarily adjacent. Statements that say an item is neither the easternmost nor the westernmost restrict it to the middle positions, which removes it as a candidate for the extreme positions and often identifies the southernmost one.

Family relationship deduction

Kinship terms follow from sex and generation. A parent's sibling is an aunt or an uncle, and the spouse of an aunt is an uncle by marriage. A mother's parent is a grandparent: a mother's mother is a grandmother and a father's father is a grandfather. Spouses belong to the same generation, and the children of a couple are one generation below them. Relations are read from the target person backward through these links.

Solve such problems in steps. First, identify the spouse pairs and the parent-child links from phrases such as 'husband of' and 'mother of'. Second, identify sibling links from 'sister of' and 'brother of'. Third, place every person in a generation. Finally, read the relation between the two named people by tracing the shortest chain of links. In a chain of this kind, the sex of each person decides which term applies.

Statements that claim a person has 'only one' son or daughter must be checked against all siblings. If a person is a son of a parent and has a brother, then that parent has at least two sons, so the claim of only one son is false. The same reasoning applies to daughters, and each claim about the number of children must be tested for every sibling present in the statements.

In team or pairing puzzles, the same kinship logic is used as a set of constraints. A parent and child cannot be placed on the same team, and a limit on the number of parent-child pairs per match removes many arrangements. Assign the child-parent links first, test each candidate against every rule, and then determine who has a parent present among the participants. A person may have no parent in the tournament if neither of their parents is among the participants.

Culprit identification from statements

A culprit identification puzzle is a truth-teller and liar problem. Several people make statements, some of them are lies, and the solver must decide who performed the act from the pattern of true and false claims. The method is to find a fixed point first, such as a speaker whose reliability is given as a fact, and then derive the truth value of every other statement from it.

Negation is the main tool. The denial of 'X did it' is 'X did not do it', and the denial of 'only one of us is telling the truth' is 'at least two of us are telling the truth', not 'none of us is telling the truth'. Students often make this last error.

When a puzzle states that exactly one person is truthful, that count is itself a claim to be checked. If the reliable speaker's statement concerns how many are truthful, every other speaker must be lying, and each lie becomes a definite fact about who did or did not act. Combine these facts, confirm that exactly one culprit remains, and check every statement against the assigned truth values. A candidate that fits some statements but breaks another is rejected.

If a speaker is known to be lying and says 'Y did it', then Y did not do it. The lie itself is therefore usable evidence, and it is often the decisive step.

• culprit — the person responsible for a wrong act. 예) The culprit was identified by comparing the witness statements.
• truth-teller / liar — a speaker who always tells the truth / always lies. 예) The guard is a liar, so his claim must be false.
• deny — to state that something is not the case. 예) She denied that she had taken the file.
• consistent with — agreeing with every given fact without contradiction. 예) The assignment is consistent with all five statements.

Voter eligibility after registration deadline

Eligibility questions test how a condition restricts a group. A rule such as 'students who have successfully completed registration are eligible' states a necessary condition: every eligible student completed registration. By contraposition, a student who did not complete registration cannot be eligible. This is the deduction the premise forces.

If a whole department failed the condition, none of its members is eligible, so every eligible student belongs to some other department. That is a conclusion about the eligible group. The reverse claim, that all ineligible students belong to the failing department, does not follow, because students of other departments may also be ineligible for reasons the passage does not mention.

To be inferred 'with certainty', a statement must be true in every situation that satisfies the premise. A statement about something the premise is silent on, such as whether students of other departments registered on time, is not inferred, however plausible it sounds. Only the statement that is forced by the premise is accepted.

• eligible — meeting the conditions needed for something. 예) Only members who renew their card are eligible to borrow books.
• necessary condition — a requirement that must be met for something to hold. 예) Renewal is a necessary condition for borrowing.
• contrapositive — the statement 'if A then B' is equivalent to 'if not B then not A'. 예) If a member has not renewed, the member cannot borrow.
• with certainty — true in every case allowed by the premise, not merely possible. 예) We can conclude with certainty only what the premise forces.