Logical Connectives - Intermediate-Advanced Level: propositional connectives INTERMEDIATE-ADVANCED

Step 1: Break down the compound expression
Expression: ¬(p ∧ q)

Step 2: Evaluate inner expression first
p ∧ q = True ∧ False = False

Step 3: Apply negation
¬(False) = True
Negation reverses the truth value

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 ∧ q)

Step 3: Test all possible combinations
A statement cannot be both true and false

This is valid: If A ⊆ B and B ⊆ C, then A ⊆ C. All dogs (A) are mammals (B), all mammals (B) are animals (C), so all dogs (A) are animals (C).

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: Modus Ponens

This is Modus Ponens: If P → Q and P are true, then Q must be true.

Conclusion: This argument is VALID.

'Unless P, then Q' means ¬P → Q. If you don't study, then you fail.

Answer: ¬s → f

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 ∧ ¬p

Step 3: Test all possible combinations
Truth table:
p=T: T ∧ F = F
p=F: F ∧ T = F
Result: Always False → CONTRADICTION
This violates the Law of Non-Contradiction

Valid: Some A are B, all B are C → Some A are C.

Subset means IF element is in A, THEN it must be in B, which is implication.

Answer: x ∈ A → x ∈ B

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 = True

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

Step 1: Understand 'neither...nor' statements
'Neither p nor q' means 'not p AND not q'
Logical form: ¬p ∧ ¬q

Step 2: Build the truth table
¬p ∧ ¬q is True ONLY when both p and q are False
Truth table:
p=T, q=T → ¬T ∧ ¬T = F ∧ F = F
p=T, q=F → ¬T ∧ ¬F = F ∧ T = F
p=F, q=T → ¬F ∧ ¬T = T ∧ F = F
p=F, q=F → ¬F ∧ ¬F = T ∧ T = T

Step 3: Apply given values
p = T, q = F
¬p = F, ¬q = T
¬p ∧ ¬q = F

Answer: False

Step 1: Break down the compound expression
Expression: (p ∨ q) → r

Step 2: Evaluate inner expression first
p ∨ q = True ∨ True = True

Step 3: Evaluate outer expression
(True) → True = True
Implication is False only when antecedent is True and consequent is False

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 = False

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

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: The alarm rings only if there is an intruder
A: Alarm rings, I: There is an intruder

Step 3: Convert to logical form
Logical form: A → I
Equivalent: If the alarm rings, then there is an intruder

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

Argument form: Modus Tollens

This is Modus Tollens: If P → Q and ¬Q are true, then ¬P must be true.

Conclusion: This argument is VALID.

Valid: Some A are B, no B are C → Some A are not C.

Step 1: Understand necessary and sufficient conditions
• P is NECESSARY for Q: Q cannot be true without P (Q → P)
• P is SUFFICIENT for Q: P being true guarantees Q (P → Q)
• P is BOTH: P if and only if Q (P ↔ Q)

Step 2: Analyze the relationship
P: Being a square
Q: Being a rectangle

Step 3: Determine the condition type
All squares are rectangles (sufficient), but not all rectangles are squares (not necessary)

Answer: Sufficient but not necessary

Step 1: Understand 'neither...nor' statements
'Neither p nor q' means 'not p AND not q'
Logical form: ¬p ∧ ¬q

Step 2: Build the truth table
¬p ∧ ¬q is True ONLY when both p and q are False
Truth table:
p=T, q=T → ¬T ∧ ¬T = F ∧ F = F
p=T, q=F → ¬T ∧ ¬F = F ∧ T = F
p=F, q=T → ¬F ∧ ¬T = T ∧ F = F
p=F, q=F → ¬F ∧ ¬F = T ∧ T = T

Step 3: Apply given values
p = T, q = T
¬p = F, ¬q = F
¬p ∧ ¬q = F

Answer: False

Step 1: Understand necessary and sufficient conditions
• P is NECESSARY for Q: Q cannot be true without P (Q → P)
• P is SUFFICIENT for Q: P being true guarantees Q (P → Q)
• P is BOTH: P if and only if Q (P ↔ Q)

Step 2: Analyze the relationship
P: Being a square
Q: Being a rectangle

Step 3: Determine the condition type
All squares are rectangles (sufficient), but not all rectangles are squares (not necessary)

Answer: Sufficient but not necessary