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Proof step 37 from proof sortedchars-no-dups (previous step)

⊦ filter ('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 38:
⊦ filter(chars, chars0) = [] ↔ (∀ char. char ∈ chars → ¬ char ∈ chars0)
Proof step 39:
⊦ ∀ char. char ∈ 'j' ' + 'k' ' + 'l' ' + 'm' ' + 'n' ' + 'o' ' + 'p' ' + 'q' ' + 'r' ' + 's' ' → ¬ char ∈ '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] → [ 't' ' + 'u' ' + 'v' ' + 'w' ' + 'x' ' + 'y' ' + 'z' ' + '|' ' + '~' ', 'j' ' + 'k' ' + 'l' ' + 'm' ' + 'n' ' + 'o' ' + 'p' ' + 'q' ' + 'r' ' + 's' '] flags: rewrite

Additional information

Used terms: 't' ' + 'u' ' + 'v' ' + 'w' ' + 'x' ' + 'y' ' + 'z' ' + '|' ' + '~' ', 'j' ' + 'k' ' + 'l' ' + 'm' ' + 'n' ' + 'o' ' + 'p' ' + 'q' ' + 'r' ' + 's' '