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Proof step 114 from proof n2s-s2n-identity (previous step)

m ≠ 0, m < 4, ∀ m1. m1 < (((((m + 1) + 1) + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ pos(('7' ' + '8' ' + '9' ')[m -1], '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 115:
m < 4, ∀ m1. m1 < (((((m + 1) + 1) + 1) + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ m = 0, pos(('7' ' + '8' ' + '9' ')[m - 1], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 6 + m

Rule argument

Additional information

Used simplifier rules: pred-03: ⊦ n -1 = n - 1; from specification nat