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Proof step 37 from proof natstring-chars (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
m0 < 5, ∀ m1, char. m1 < ((((m0 + 1) + 1) + 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' ⊦ ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '0', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '9', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '8', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '7', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '6', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '5', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '4', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '3', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '2', ('5' ' + '6' ' + '7' ' + '8' ' + '9' ')[m0] = '1'

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;