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Proof step 59 from proof n2s-s2n-identity (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
∀ m1. m1 < (((0 + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ pos('4', '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 4

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]