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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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c○: Michael Kohlhase 188<br />

The entailment theorem has a syntactic counterpart for some calculi. This result shows a close<br />

connection between the derivability relation and the implication connective. Again, the two should<br />

not be confused, even though this time, both are syntactic.<br />

The main idea in the following proof is to generalize the inductive hypothesis from proving A ⇒ B<br />

to proving A ⇒ C, where C is a step in the proof of B. The assertion is a special case then, since<br />

B is the last step in the proof of B.<br />

The Deduction Theorem<br />

Theorem 321 If H, A ⊢ B, then H ⊢ A ⇒ B<br />

Proof: By induction on the proof length<br />

P.1 Let C1, . . . , Cm be a proof of B from the hypotheses H.<br />

P.2 We generalize the induction hypothesis: For all l (1 ≤ i ≤ m) we construct proofs<br />

H ⊢ A ⇒ Ci. (get A ⇒ B for i = m)<br />

P.3 We have to consider three cases<br />

P.3.1 Case 1: Ci axiom or Ci ∈ H:<br />

P.3.1.1 Then H ⊢ Ci by construction and H ⊢ Ci ⇒ A ⇒ Ci by Subst from Axiom 1.<br />

P.3.1.2 So H ⊢ A ⇒ Ci by MP.<br />

P.3.2 Case 2: Ci = A:<br />

P.3.2.1 We have already proven ∅ ⊢ A ⇒ A, so in particular H ⊢ A ⇒ Ci.<br />

(more hypotheses do not hurt)<br />

P.3.3 Case 3: everything else:<br />

P.3.3.1 Ci is inferred by MP from Cj and Ck = Cj ⇒ Ci for j, k < i<br />

P.3.3.2 We have H ⊢ A ⇒ Cj and H ⊢ A ⇒ Cj ⇒ Ci by IH<br />

P.3.3.3 Furthermore, (A ⇒ Cj ⇒ Ci) ⇒ (A ⇒ Cj) ⇒ A ⇒ Ci by Axiom 2 and Subst<br />

P.3.3.4 and thus H ⊢ A ⇒ Ci by MP (twice).<br />

P.4 We have treated all cases, and thus proven H ⊢ A ⇒ Ci for (1 ≤ i ≤ m).<br />

P.5 Note that Cm = B, so we have in particular proven H ⊢ A ⇒ B.<br />

c○: Michael Kohlhase 189<br />

In fact (you have probably already spotted this), this proof is not correct. We did not cover all<br />

cases: there are proofs that end in an application of the Subst rule. This is a common situation,<br />

we think we have a very elegant and convincing proof, but upon a closer look it turns out that<br />

there is a gap, which we still have to bridge.<br />

This is what we attempt to do now. The first attempt to prove the subst case below seems to<br />

work at first, until we notice that the substitution [B/X] would have to be applied to A as well,<br />

which ruins our assertion.<br />

The missing Subst case<br />

Oooops: The proof of the deduction theorem was incomplete<br />

(we did not treat the Subst case)<br />

100

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