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Proof step 39 from proof sortedchars-no-dups (previous step)

⊦ ∀ char. char ∈ 'j' ' + 'k' ' + 'l' ' + 'm' ' + 'n' ' + 'o' ' + 'p' ' + 'q' ' + 'r' ' + 's' ' → ¬ char ∈ 't' ' + 'u' ' + 'v' ' + 'w' ' + 'x' ' + 'y' ' + 'z' ' + '|' ' + '~' '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
⊦ char ≠ 't' ∧ char ≠ 'u' ∧ char ≠ 'v' ∧ char ≠ 'w' ∧ char ≠ 'x' ∧ char ≠ 'y' ∧ char ≠ 'z' ∧ char ≠ '|' ∧ char ≠ '~', char ≠ 'j' ∧ char ≠ 'k' ∧ char ≠ 'l' ∧ char ≠ 'm' ∧ char ≠ 'n' ∧ char ≠ 'o' ∧ char ≠ 'p' ∧ char ≠ 'q' ∧ char ≠ 'r' ∧ char ≠ 's'

Rule argument

Additional information

Used simplifier rules: ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ char ∈ char0 ' ↔ char = char0;% ⊦ char ∈ chars + chars0 ↔ ¬ ¬(char ∈ chars ∨ char ∈ chars0);