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Proof step 32 from proof n2s-chars (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 33:
m = m0 + 1, m < 6, ∀ m1, char. m1 < (((m + 1) + 1) + 1) + 1 → true → ¬ char ∈ '0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9') → ¬ char ∈ s→l(n→s(m1)) ⊦ m = 0, ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '1', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '2', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '3', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '4', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '5', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '6', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '7', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '8', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '9', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '0'

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0