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Proof step 35 from proof sortedchars-no-dups (previous step)

dups(s→l("tuvwxyz|~")) ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: s2l-rec-01: str ≠ "" ⊦ s→l((* str)) = str.char1 ' + s→l(str.rstring);%s2l-empty: ⊦ s→l("") = [];% ⊦ ¬ dups(char ');% ⊦ char ∈ char0 ' ↔ char = char0;% ⊦ char ∈ chars + chars0 ↔ ¬ ¬(char ∈ chars ∨ char ∈ chars0);% ⊦ dups(char ' + chars) ↔ ¬ ¬(char ∈ chars ∨ dups(chars));% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ chars + [] = chars;