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Proof step 41 from proof n2s-s2n-identity (previous step)

m < 8, ∀ m1. m1 < (m + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ m = 0, pos(('3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m - 1], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 2 + m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 42:
m = m0 + 1, m < 8, ∀ m1. m1 < (m + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ m = 0, pos(('3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 2 + m

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0