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Proof step 51 from proof sortedchars-no-dups (previous step)

⊦ filter ('[' ' + ']' ' + '^' ' + '_' ' + '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' ' + '|' ' + '~' ') = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 52:
⊦ filter(chars, chars0) = [] ↔ (∀ char. char ∈ chars → ¬ char ∈ chars0)
Proof step 53:
⊦ ∀ char. char ∈ '[' ' + ']' ' + '^' ' + '_' ' + 'a' ' + 'b' ' + 'c' ' + 'd' ' + 'e' ' + 'f' ' + 'g' ' + 'h' ' + 'i' ' → ¬ char ∈ ('j' ' + 'k' ' + 'l' ' + 'm' ' + 'n' ' + 'o' ' + 'p' ' + 'q' ' + 'r' ' + 's' ') + 't' ' + 'u' ' + 'v' ' + 'w' ' + 'x' ' + 'y' ' + 'z' ' + '|' ' + '~' '

Rule argument

spec: list-set inst: charlist name: filter-empty-def seq: ⊦ filter(chars, chars0) = [] ↔ (∀ char. char ∈ chars → ¬ char ∈ chars0) subst: [chars0, chars] → [ ( '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' '] flags: rewrite

Additional information

Used terms: ( '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' '