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# lecture18 - 1-9 CP 11(P → Q v(Q → P 9 MI Theorems and...

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Theorems Theorems are statements that can be proved without any premises. Theorems are tautologies (and conversely). They are true just in virtue of their logical form. Thus any argument that has a theorem as its conclusion will be valid. Example: Grass is green, so either the sky is blue or it’s not. Similarly, an ‘argument’ with a theorem as its conclusion and no premises will be valid. CP and RAA allow us to derive theorems from no premises, since they allow us to introduce assumptions.

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Theorems Example: .˙. P → (Q →P) 1) P Assume 2) Q Assume 3) P v P 1, Add 4) P 4, Re 5) Q → P 2-4, CP 6) P → (Q → P) 1-5, CP
Theorems Here’s a more complicated example. .˙. (P → Q) v (Q → P) 1) ~ (P → Q) Assume 2) Q Assume 3) ~P Assume 4) ~(~P v Q) 1, MI 5) ~~P ● ~Q 4, DM 6) ~Q 5, Simp 7) Q ● ~Q 2, 6 Conj 8) P 3-7, RAA 9) Q → P 2-8, CP 10) ~ (P → Q) → (Q → P)

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Unformatted text preview: 1-9, CP 11) (P → Q) v (Q → P) 9, MI Theorems and Valid Arguments A connection between valid arguments and theorems: An argument of L is valid just in case its corresponding conditional is a theorem. An argument’s corresponding conditional is what we get when we form the conditional using the conjunction of an argument’s premises as the antecedent, and the conclusion as the consequent . Argument: P v Q, ~P .˙. Q Corresponding conditional: [(P v Q) ● ~P] → Q So it’s possible to show that an argument is valid by forming the corresponding conditional and then deriving it from no premises. More Examples of Theorems [(H → I) → H] → H [(P ● Q) v R] → [(~R v Q) → (P → Q)] [P v (~P ● Q)] ↔ (P v Q)...
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lecture18 - 1-9 CP 11(P → Q v(Q → P 9 MI Theorems and...

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