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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. ( )) ( ) ( )