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Proof step 12 from proof natstring-chars (previous step)

m < 10, char ∈ ('0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9'))[m] + "", m < 10, ∀ m1, char. m1 < m → char ∈ n→s(m1) → char = '0' ∨ char = '9' ∨ char = '8' ∨ char = '7' ∨ char = '6' ∨ char = '5' ∨ char = '4' ∨ char = '3' ∨ char = '2' ∨ char = '1' ⊦ char = '1', char = '2', char = '3', char = '4', char = '5', char = '6', char = '7', char = '8', char = '9', char = '0'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: char-in-one: ⊦ char ∈ char0 ' ↔ char = char0;% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ str1 + "" = str1;% ⊦ char + str = char ' + str;% ⊦ char + chars = char ' + chars;% ⊦ char + char0 = char ' + char0 ';