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Proof step 92 from proof n2s-s2n-identity (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 93:
m0 = m + 1, m0 < 5, ∀ m1. m1 < ((((m0 + 1) + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ m0 = 0, pos(('6' ' + '7' ' + '8' ' + '9' ')[m], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 5 + m0

Rule argument

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

Additional information

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