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Proof step 4 from proof natstring-chars (previous step)

char ∈ n→s(m) ⊦ char ∈ s→l("0123456789")

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
char ∈ n→s(m) ⊦ char = '0', char = '9', char = '8', char = '7', char = '6', char = '5', char = '4', char = '3', char = '2', char = '1'

Rule argument

Additional information

Used simplifier rules: ⊦ char ∈ char0 ' ↔ char = char0;% ⊦ char ∈ chars + chars0 ↔ ¬ ¬(char ∈ chars ∨ char ∈ chars0);% ⊦ ¬ char ∈ char0 ' + chars ↔ char ≠ char0 ∧ ¬ char ∈ chars;% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ chars + [] = chars;% ⊦ s→l("") = [];%str ≠ "" ⊦ s→l((* str)) = str.char1 ' + s→l(str.rstring);