Applied Proof Theory: Proof Interpretations and their Use in by Ulrich Kohlenbach

By Ulrich Kohlenbach

This is often the 1st therapy in e-book structure of proof-theoretic modifications - referred to as evidence interpretations - that makes a speciality of purposes to boring arithmetic. It covers either the mandatory logical equipment in the back of the facts interpretations which are utilized in fresh functions in addition to – through prolonged case reviews – undertaking a few of these functions in complete element. This topic has old roots within the Fifties. This ebook for the 1st time tells the entire story.

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24. There exists a primitive recursive function f0 : N → N such that there is no computable function Φ : N → N with ∀k ∈ N∃n ≤ Φ (k) n ≥ k ∧ ∀m ≥ k( f0 (n) ≤ f0 (m)) . 3 Herbrand’s theorem and the no-counterexample interpretation 29 Proof: Let e ∈ N be such that {k ∈ N : {e}(k) ↓} = {k ∈ N : ∃n ∈ N T (e, k, n)} is undecidable, where the primitive recursive Kleene T -predicate satisfies (1) ∀k, n1 , n2 ∈ N(T (e, k, n1 ) ∧ T (e, k, n2 ) → n1 = n2 ). g. 30 below). It is clear that f0 is primitive recursive.

E. in the sense of Kleene (these results are proved in [215] which provides a thorough discussion of the modus ponens problem for the no-counterexample interpretation). The reason for the weakness of the no-counterexample interpretation is the weakness of the Herbrand normal form F H of formulas F of complexity ∃x∀y∃z F0 (x, y, z) or higher (such as (A → B) pr above). Then the passage from F H to F requires AC (though only from numbers to numbers) for ∀-formulas (and beyond), which in general are undecidable.

11. 1) From a deep result of Gelfand and Schneider, stating that if a, b that are √algebraic, a = 0, 1 and b irrational, then ab is transcendental, it follows √ √2 √ √ 2 2 is transcendental and, therefore, irrational. So it is the pair ( 2 , 2) which satisfies the proposition. 9: take a := 2 and b := 2 log2 (3). b is irrational since log2 (3) = m/n for some m, n ∈ N∗ would imply that 2m = 3n which is impossible. Clearly, ab = 3 is rational. Here is another example (communicated by H. 12. (e − π is irrational) or (e + π is irrational).

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