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Unformatted text preview: kValidity vs Validity in Q : A formula of Q that is kvalid 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 V x 00 V x 000 [(F **0 x x 00 ∧ F **0 x 00 x 000 ) ⊃ F **0 x x 000 ] • In more standard notation: p ( ∀ x)( ∀ y)( ∀ z)[(Rxy ∧ Ryz) ⊃ Rxz] . q V x W x 00 F **0 x x 00 • In more standard notation: q ( ∀ x)( ∃ y)Rxy . r W x F **0 x x • In more standard notation: r ( ∃ x)Rxx . Informally, p asserts that the 2place 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 kvalid for all (finite) k , but A is not valid [ ø Q A ]. Proof. First, we will show (informally) that A is kvalid, for all k . Our (informal) argument will involve showing that A is true on all 1element interpretations, and all 2element interpretations, and ..., and all kelement interpretations, for all k . We will do this by arguing (informally) that we cannot make A false on any kelement 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 kelement interpretations of Q , for all k ....
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This note was uploaded on 08/01/2008 for the course PHIL 140A taught by Professor Fitelson during the Spring '07 term at Berkeley.
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