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General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

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The deduction theorem and the entailment theorem together allow us to understand the claim that<br />

the two formulations of soundness (A ⊢ B implies A |= B and ⊢ A implies |=B) are equivalent.<br />

Indeed, if we have A ⊢ B, then by the deduction theorem ⊢ A ⇒ B, and thus |=A ⇒ B by<br />

soundness, which gives us A |= B by the entailment theorem. The other direction and the<br />

argument for the corresponding statement about completeness are similar.<br />

Of course this is still not the version of the proof we originally wanted, since it talks about the<br />

Hilbert Calculus H 1 , but we can show that H 1 and H 0 are equivalent.<br />

But as we will see, the derivability relations induced by the two caluli are the same. So we can<br />

prove the original theorem after all.<br />

The Deduction Theorem for H 0<br />

Lemma 324 ⊢ H 1 = ⊢ H 0<br />

Proof:<br />

P.1 All H 1 axioms are H 0 theorems. (by Subst)<br />

P.2 For the other direction, we need a proof transformation argument:<br />

P.3 We can replace an application of MP followed by Subst by two Subst applications followed<br />

by one MP.<br />

P.4 . . . A ⇒ B . . . A . . . B . . . [C/X](B) . . . is replaced by<br />

. . . A ⇒ B . . . [C/X](A) ⇒ [C/X](B) . . . A . . . [C/X](A) . . . [C/X](B) . . .<br />

P.5 Thus we can push later Subst applications to the axioms, transforming a H 0 proof into<br />

a H 1 proof.<br />

Corollary 325 H, A ⊢ H 0 B, iff H ⊢ H 0 A ⇒ B.<br />

Proof Sketch: by MP and ⊢ H 1 = ⊢ H 0<br />

c○: Michael Kohlhase 193<br />

We can now collect all the pieces and give the full statement of the soundness theorem for H 0<br />

H 0 is sound (full version)<br />

Theorem 326 For all propositions A, B, we have A ⊢ H 0 B implies A |= B.<br />

Proof:<br />

P.1 By deduction theorem A ⊢ H 0 B, iff ⊢ A ⇒ C,<br />

P.2 by the first soundness theorem this is the case, iff |=A ⇒ B,<br />

P.3 by the entailment theorem this holds, iff A |= C.<br />

2.6.5 A Calculus for Mathtalk<br />

c○: Michael Kohlhase 194<br />

In our introduction to Subsection 2.6.0 we have positioned Boolean expressions (and proposition<br />

logic) as a system for understanding the mathematical language “mathtalk” introduced in Subsection<br />

2.2.1. We have been using this language to state properties of objects and prove them all<br />

through this course without making the rules the govern this activity fully explicit. We will rectify<br />

this now: First we give a calculus that tries to mimic the the informal rules mathematicians use<br />

102

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