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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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;