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Proof step 76 from proof n2s-s2n-identity (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 77:
∀ m1. m1 < ((((0 + 1) + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ pos('5', '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 5

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1; from specification list[charlist]%get-zero-01: ⊦ (char ' + chars)[0] = char; from specification list-del[charlist]