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Proof step 86 from proof sortedchars-no-dups (previous step)

filter(s→l("ABCDEFGHIJKLM"), s→l("NOPQRSTUVWXYZ") + s→l("[]^_abcdefghi") + s→l("jklmnopqrs") + s→l("tuvwxyz|~")) ≠ [] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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