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Proof step 54 from proof n2s-s2n-identity (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 55:
m0 < 7, ∀ m1. m1 < ((m0 + 1) + 1) + 1 → true → s→n(n→s(m1)) = m1 ⊦ m0 = 0, pos(('4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0 - 1], '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ') = 3 + m0

Rule argument

Additional information

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