Logical Connectives - Intermediate-Advanced Level: logical equivalence INTERMEDIATE-ADVANCED

Complement means element is NOT in the set, which is logical NOT.

Answer: ¬(x ∈ A)

Test cases:
- A knight → statement true: 'knight or B knave' → true (first part true) ✓. B says 'A knave' which is false, so B knave ✓.
- A knave → statement false: 'knight or B knave' false → both parts false → 'knight' false (ok), 'B knave' false → B knight. Then B says 'A knave' which would be true (since A knave), but B knight must tell truth ✓. This also works? Wait, if A knave and B knight: A's statement 'knight or B knave' = 'false or false' = false (knave lies) ✓. B's statement 'A knave' = true (knight truth) ✓. Two solutions? But puzzle assumes unique. Check carefully: With A knave, B knight: A says 'knight or B knave' = 'false or false' = false (good lie). B says 'A knave' = true (good truth). Both work. So puzzle ambiguous. We'll take first solution: A knight, B knave.

Implication is not symmetric. Converse is not equivalent to original.

p=True, q=False: p→q=False, q→p=True → Different!

Step 1: Understand 'only if' statements
'P only if Q' means 'If P, then Q' (P → Q)
This is DIFFERENT from 'If Q then P' (Q → P)

Key insight: 'only if' introduces a NECESSARY condition
Q is necessary for P (P cannot be true without Q)

Step 2: Identify components
Statement: You will pass only if you study
P: You pass, S: You study

Step 3: Convert to logical form
Logical form: P → S
Equivalent: If you pass, then you studied

Step 4: Important distinction
Note: Studying is necessary but not sufficient
'Only if' ≠ 'If and only if'
'Only if' gives one direction only (→)
'If and only if' gives both directions (↔)

Step 1: Understand the classifications
• Tautology: Always True for all possible truth values
• Contradiction: Always False for all possible truth values
• Contingent: True for some values, False for others

Step 2: Analyze the expression
Expression: (p → q) → p

Step 3: Test all possible combinations
This is contingent - depends on values of p and q

Step 1: Understand the conjunction (AND) operator
The conjunction p ∧ q is True ONLY when BOTH p and q are True.

Step 2: Apply the given values
p = True, q = False

Step 3: Evaluate p ∧ q
Since at least one of p or q is False, p ∧ q = False

Step 1: Understand what logical equivalence means
Two expressions are logically equivalent if they have the same truth value for ALL possible combinations of variables.

Step 2: Analyze the expressions
Expression 1: p → q
Expression 2: ¬q → ¬p

Step 3: Apply Contrapositive equivalence
A conditional statement and its contrapositive are ALWAYS equivalent.
If p → q, then ¬q → ¬p is also true.
These expressions ARE equivalent.

Step 1: Understand 'unless' statements
'P unless Q' means 'If not Q, then P'
In logical form: 'P unless Q' ≡ '¬Q → P'

Step 2: Identify components
Original: You will fail unless you study
p: You will fail, q: You study

Step 3: Convert to if-then form
'unless' tells us what happens if the condition is NOT met
Logical form: ¬q → p

Step 4: Write in English
Equivalent statement: If you do not study, then you will fail

Step 1: Understand the disjunction (OR) operator
The disjunction p ∨ q is True when AT LEAST ONE of p or q is True.
It is False ONLY when both p and q are False.

Step 2: Apply the given values
p = True, q = False

Step 3: Evaluate p ∨ q
Since at least one of p or q is True, p ∨ q = True

Rule: Simplification (∧-elimination)

From a conjunction P ∧ Q, you can derive P (or Q) alone.

Answer: p

Step 1: Understand the conjunction (AND) operator
The conjunction p ∧ q is True ONLY when BOTH p and q are True.

Step 2: Apply the given values
p = False, q = True

Step 3: Evaluate p ∧ q
Since at least one of p or q is False, p ∧ q = False

Argument form: Denying the Antecedent

Fallacy! You might still pass without studying (natural talent, cheating, easy exam). ¬P does NOT imply ¬Q.

Conclusion: This argument is INVALID.

Step 1: Break down the expression
Expression: ¬p ∨ (q ∧ r)
Given: p=T, q=T, r=F

Step 2: Evaluate inner expressions first
¬p = F
q ∧ r = T ∧ F = F

Step 3: Evaluate outer expression
F ∨ F = False
Remember: OR is True when at least one operand is True

Step 1: Understand the conjunction (AND) operator
The conjunction p ∧ q is True ONLY when BOTH p and q are True.

Step 2: Apply the given values
p = True, q = False

Step 3: Evaluate p ∧ q
Since at least one of p or q is False, p ∧ q = False

Rule: Disjunctive Syllogism

If P ∨ Q is true and P is false, then Q must be true.

Answer: q

'Either P or Q, but not both' is exclusive OR: P ⊕ Q.

Answer: p ⊕ q

Rule: Disjunction Introduction (∨-intro)

If P is true, then P ∨ Q is true for any Q (addition rule).

Answer: p ∨ q

Step 1: Understand the implication (→) operator
The implication p → q is False ONLY when p is True but q is False.
In all other cases, it is True.

Truth table for p → q:
p=T, q=T → Result=T
p=T, q=F → Result=F (the only False case)
p=F, q=T → Result=T
p=F, q=F → Result=T

Step 2: Apply the given values
p = False, q = True

Step 3: Evaluate p → q
Since this is not the case where p is True and q is False, p → q = True

False dilemma (either/or fallacy). Presents only two options when more exist (e.g., you could disagree but still love your country).

Case analysis: If A knight → B knave (A's truth) → A and C different (B's lie) → C knight? But C says 'A is knight' which would be true, consistent. Wait, need full check.

Actually solve: Assume A knight → 'B knave' true → B knave → B's statement 'A and C same' is false → A and C different → C knave → C says 'A knight' which is false (since A knight?) Contradiction.

Therefore A knave → 'B knave' false → B knight → B's statement true → A and C same → C knave → C says 'A knight' false (since A knave) ✓. Solution: A knave, B knight, C knave.