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Proof step 43 from proof n2s-chars (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
m0 < 3, ∀ m1, char. m1 < ((((((m0 + 1) + 1) + 1) + 1) + 1) + 1) + 1 → true → ¬ char ∈ '0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9') → ¬ char ∈ s→l(n→s(m1)) ⊦ ('7' ' + '8' ' + '9' ')[m0] = '1', ('7' ' + '8' ' + '9' ')[m0] = '2', ('7' ' + '8' ' + '9' ')[m0] = '3', ('7' ' + '8' ' + '9' ')[m0] = '4', ('7' ' + '8' ' + '9' ')[m0] = '5', ('7' ' + '8' ' + '9' ')[m0] = '6', ('7' ' + '8' ' + '9' ')[m0] = '7', ('7' ' + '8' ' + '9' ')[m0] = '8', ('7' ' + '8' ' + '9' ')[m0] = '9', ('7' ' + '8' ' + '9' ')[m0] = '0'

Rule argument

Additional information

Used simplifier rules: n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;