kvalidity

# kvalidity - k-Validity vs Validity in Q A formula of Q that...

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k -Validity vs Validity in Q : A formula of Q that is k -valid for all k , but not valid Branden Fitelson 03/13/07 Consider the following three formulas of Q [where, as always, p q (p ⊃ ∼ q) , and W p V p ]: p V x 0 V x 00 V x 000 [(F **0 x 0 x 00 F **0 x 00 x 000 ) F **0 x 0 x 000 ] In more standard notation: p ( x)( y)( z)[(Rxy Ryz) Rxz] . q V x 0 W x 00 F **0 x 0 x 00 In more standard notation: q ( x)( y)Rxy . r W x 0 F **0 x 0 x 0 In more standard notation: r ( x)Rxx . Informally, p asserts that the 2-place relation F **0 ( R ) is transitive , q asserts that F **0 ( R ) is serial , and r asserts that there is some object that bears the relation F **0 ( R ) to itself . Now, consider the following complex statement, constructed out of p , q , and r : A (p q) r Claim A asserts that if F **0 ( R ) is transitive and serial, then some object bears F **0 ( R ) to itself . Fact . A is k -valid for all (finite) k , but A is not valid [ ø Q A ]. Proof. First, we will show (informally) that A is k -valid, for all k . Our (informal) argument will involve showing that A is true on all 1-element interpretations, and all 2-element interpretations, and . . . , and all k -element interpretations, for all k . We will do this by arguing (informally) that we cannot make A false on any k -element interpretation. And, since A is a closed formula, it must either be true or false on each interpretation of Q . Thus, it will follow that A is true on all k -element interpretations of Q , for all k . Let’s think about 1-element interpretations I 1 first. In order to make A false on any interpretation I , we would need to make p and q both true, and r false on I

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