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Proof step 107 from proof sortedchars-no-dups (previous step)

⊦ ∀ char. char ∈ '.' ' + '/' ' + '0' ' + '1' ' + '2' ' + '3' ' + '4' ' + '5' ' + '6' ' + '7' ' + '8' ' + '9' ' + ':' ' + '<' ' + '=' ' + '>' ' + '?' ' + '@' ' → ¬ char ∈ ('A' ' + 'B' ' + 'C' ' + 'D' ' + 'E' ' + 'F' ' + 'G' ' + 'H' ' + 'I' ' + 'J' ' + 'K' ' + 'L' ' + 'M' ') + ('N' ' + 'O' ' + 'P' ' + 'Q' ' + 'R' ' + 'S' ' + 'T' ' + 'U' ' + 'V' ' + 'W' ' + 'X' ' + 'Y' ' + 'Z' ') + ('[' ' + ']' ' + '^' ' + '_' ' + 'a' ' + 'b' ' + 'c' ' + 'd' ' + 'e' ' + 'f' ' + 'g' ' + 'h' ' + 'i' ') + ('j' ' + 'k' ' + 'l' ' + 'm' ' + 'n' ' + 'o' ' + 'p' ' + 'q' ' + 'r' ' + 's' ') + 't' ' + 'u' ' + 'v' ' + 'w' ' + 'x' ' + 'y' ' + 'z' ' + '|' ' + '~' '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 108:
⊦ char ≠ 'A' ∧ char ≠ 'B' ∧ char ≠ 'C' ∧ char ≠ 'D' ∧ char ≠ 'E' ∧ char ≠ 'F' ∧ char ≠ 'G' ∧ char ≠ 'H' ∧ char ≠ 'I' ∧ char ≠ 'J' ∧ char ≠ 'K' ∧ char ≠ 'L' ∧ char ≠ 'M' ∧ char ≠ 'N' ∧ char ≠ 'O' ∧ char ≠ 'P' ∧ char ≠ 'Q' ∧ char ≠ 'R' ∧ char ≠ 'S' ∧ char ≠ 'T' ∧ char ≠ 'U' ∧ char ≠ 'V' ∧ char ≠ 'W' ∧ char ≠ 'X' ∧ char ≠ 'Y' ∧ char ≠ 'Z' ∧ char ≠ '[' ∧ char ≠ ']' ∧ char ≠ '^' ∧ char ≠ '_' ∧ char ≠ 'a' ∧ char ≠ 'b' ∧ char ≠ 'c' ∧ char ≠ 'd' ∧ char ≠ 'e' ∧ char ≠ 'f' ∧ char ≠ 'g' ∧ char ≠ 'h' ∧ char ≠ 'i' ∧ char ≠ 'j' ∧ char ≠ 'k' ∧ char ≠ 'l' ∧ char ≠ 'm' ∧ char ≠ 'n' ∧ char ≠ 'o' ∧ char ≠ 'p' ∧ char ≠ 'q' ∧ char ≠ 'r' ∧ char ≠ 's' ∧ char ≠ 't' ∧ char ≠ 'u' ∧ char ≠ 'v' ∧ char ≠ 'w' ∧ char ≠ 'x' ∧ char ≠ 'y' ∧ char ≠ 'z' ∧ char ≠ '|' ∧ char ≠ '~', char ≠ '.' ∧ char ≠ '/' ∧ char ≠ '0' ∧ char ≠ '1' ∧ char ≠ '2' ∧ char ≠ '3' ∧ char ≠ '4' ∧ char ≠ '5' ∧ char ≠ '6' ∧ char ≠ '7' ∧ char ≠ '8' ∧ char ≠ '9' ∧ char ≠ ':' ∧ char ≠ '<' ∧ char ≠ '=' ∧ char ≠ '>' ∧ char ≠ '?' ∧ char ≠ '@'

Rule argument

Additional information

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