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Proof step 2 from proof e-01 (previous step)

dups([]) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ dups(x) ↔ (∃ a, x0, y, z. x = x0 + (a + (y + (a + z))))

Rule argument

spec: list-dup inst: name: dups seq: ⊦ dups(x) ↔ (∃ a, x0, y, z. x = x0 + (a + (y + (a + z)))) subst: [x] → [[]] flags: rewrite, current

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list Used terms: []