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Proof step 8 from proof n2s-chars (previous step)

m < 10, char ∈ s→l(('0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9'))[m] + ""), 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: ⊦ char ∈ char0 ' ↔ char = char0;% ⊦ s→l(char ') = char ';% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ str1 + "" = str1;% ⊦ char + str = char ' + str;% ⊦ char + chars = char ' + chars;% ⊦ char + char0 = char ' + char0 ';