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Proof step 15 from proof rec (previous step)

⊦ dups(a ' + x) ↔ (∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z), a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ dups(x) ↔ (∃ a, x0, y, z. x = x0 + (a + (y + (a + z))))
Proof step 17:
⊦ (∃ a0, x0, y, z. a ' + x = x0 + (a0 + (y + (a0 + z)))) ↔ (∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z), a ∈ x

Rule argument

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

Additional information

Used terms: a ' + x