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Proof step 51 from proof n2s-chars (previous step)

10 ≤ m, char ∈ s→l(n→s(m / 10) + n→s(m % 10)), ∀ m1, char. m1 < m → true → ¬ char ∈ '0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9') → ¬ char ∈ s→l(n→s(m1)) ⊦ char = '0', char = '9', char = '8', char = '7', char = '6', char = '5', char = '4', char = '3', char = '2', char = '1'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ char ∈ chars + chars0 ↔ ¬ ¬(char ∈ chars ∨ char ∈ chars0);% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ s→l(str + str0) = s→l(str) + s→l(str0);