Last Update: 3 February 2009
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Note: In what follows,
I.e., the universal quantifier distributes over conjunction, but not disjunction, and the existential quantifier distributes over disjunction, but not conjunction.
proof:
Then: tval(LHS) = T, but tval(RHS) = F
Then: tval(LHS) = F, but tval(RHS) = T
Show: LHS & RHS always have same tval:
Case 1: Suppose tval(LHS) = T, & show tval(RHS) = T:
That is, tval(P(a) ∨ Q(a)) = T.
Case 1a: Suppose tval(P(a)) = T
Case 1b: Suppose tval(Q(a)) = T
So, in either case, the tval(RHS) of the original equivalence = T
Case 2: Suppose tval(RHS) = T & show tval(LHS) = T:
So, either tval(∃xP(x)) = T or tval(∃xQ(x)) = T.
If tval(∃xP(x)) = T, there's something in the domain that is P; call it "a". So: tval(P(a)) = T
If tval(∃xQ(x)) = T, there's something in the domain that is Q; call it "b". So: tval(Q(b)) = T
Case 2a: Suppose tval(P(a)) = T
Case 2b: Suppose tval(Q(b)) = T
So, in either case, the tval(LHS) of the original equivalence = T
QED.