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Proof step 116 from proof n2s-s2n-identity (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 117:
m0 < 3, ∀ m1. m1 < ((((((m0 + 1) + 1) + 1) + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ pos(('7' ' + '8' ' + '9' ')[m0], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 7 + m0

Rule argument

Additional information

Used simplifier rules: l-num-05: n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2