Download Though the universal quantifier distributes over conjunction, it does

Though the universal quantifier distributes over conjunction, it does not distribute over disjunction
( ( )
Prove:
( ))
( )
( )
where != means “is not logically equivalent to”
By contradiction.
Assume:
Technique: Since the assumption asserts the distributional properties of the universal quantifier for all
possible instantiations of p(X) and q(X), finding a single example for which the assumption does not hold
is sufficient to prove the theorem.
Let the domain be Z, the set of integers.
Let tval(S) be a function that maps to {T/F} depending on the truth value of the predicate calculus
sentence S
Let p(X) be a function that maps to T if X is even, F otherwise
Let q(x) be a function that maps to T is X is odd, F otherwise
Observation: tval(LHS) is T, because any integer is either even or odd. Example: suppose X = 3. Then
p(3) is F and q(3) is T. T F, of course, is T by the axiom of disjunction.
But tval(RHS) is F because the two disjuncts are F:


( ) is clearly F, since not all integers are even
( ) is clearly F, since not all integers are odd
Therefore, the assumption that
is false. It’s opposite, therefore is true and
( ( )
which is what we wanted to prove.
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