Logical Connectives - Intermediate Level: conjunction INTERMEDIATE

∧ doesn't distribute over ∨ that way. Correct: p∧(q∨r) ≡ (p∧q) ∨ (p∧r)

p=False, q=True, r=False: Left=False, Right=True → Different!

Valid: Some A are B, all B are C → Some A are 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 = False

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

Step 1: Understand the original statement
Original: p → q means "If the alarm rings, then I wake up"

Step 2: Understand Contrapositive
Contrapositive switches AND negates both parts: ¬q → ¬p
If the original is p → q, the contrapositive is ¬q → ¬p
Important: A conditional and its contrapositive are LOGICALLY EQUIVALENT

Step 3: Apply to our statement
Contrapositive: If I wake up is false, then the alarm rings is false

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

Answer: ¬(x ∈ A)

Valid: No A are B, all C are A → No C are B.

This is valid in form, but the premise 'All birds can fly' is false. Validity is about logical structure, not factual truth. Form: All A are B, C is A → C is B.

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) ∨ False = True
Since OR is True when at least one operand is 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: 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 (↔)

Rule: Disjunction Introduction (∨-intro)

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

Answer: p ∨ q

Step 1: Analyze the statements
A says: 'B is a knave'
B says: 'We are different types'

Step 2: Test Case 1 - Both knights
If both are knights, they both tell truth.
A (truth): 'B is a knave' - but B is a knight, so FALSE ✗
This case fails.

Step 3: Test Case 2 - A knight, B knave
A (truth): 'B is a knave' - TRUE ✓
B (lie): 'We are different types' - TRUE but B must lie ✗
This case fails.

Step 4: Test Case 3 - A knave, B knight
A (lie): 'B is a knave' - but B is knight, so this is FALSE, which means A is lying correctly ✓
B (truth): 'We are different types' - TRUE ✓
This case works!

Wait, let me recalculate...
Actually if A is knave lying that 'B is a knave', then B is actually a knight (correct).
B is knight saying 'We are different' is TRUE.

Answer: A is a knave, B is a knight

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: Break down the expression
Expression: p → (q ∨ r)
Given: p=F, q=T, r=T

Step 2: Evaluate inner expressions first
q ∨ r = T ∨ T = T

Step 3: Evaluate outer expression
p → (T) = F → T = True
Remember: Implication is False only when p is True and consequent is False

If A knave → statement 'at least one knave' true → but knave can't tell truth → impossible. So A knight → statement true → at least one knave → since A knight, B must be knave.

Rule: Simplification (∧-elimination)

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

Answer: p

Equal sets mean element is in A IF AND ONLY IF it is in B, which is biconditional.

Answer: x ∈ A ↔ x ∈ B

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

Answer: ¬s → f

Step 1: Test if A is a knight
If A is a knight, A tells the truth.
But A says 'I am a knave', which would be a lie.
Contradiction! A cannot be a knight.

Step 2: Test if A is a knave
If A is a knave, A lies.
But A says 'I am a knave', which would be true.
Contradiction! A cannot be a knave.

Step 3: Conclusion
Neither possibility works.
This statement is a LOGICAL PARADOX.
No one can truthfully or falsely claim to be a knave.

Answer: This statement is impossible

Step 1: Test if A is a knight
If A is a knight, A tells the truth.
But A says 'I am a knave', which would be a lie.
Contradiction! A cannot be a knight.

Step 2: Test if A is a knave
If A is a knave, A lies.
But A says 'I am a knave', which would be true.
Contradiction! A cannot be a knave.

Step 3: Conclusion
Neither possibility works.
This statement is a LOGICAL PARADOX.
No one can truthfully or falsely claim to be a knave.

Answer: This statement is impossible

Argumentum ad populum: Popular belief doesn't make something true.