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Proof step 8 from proof dups-firstoc-rew (previous step)

a0 ∈ y, a ∈ y ⊦ dups(y), ∃ y0, a1, z0. a ' + y + a0 ' + z = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ x ↔ (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y)
Proof step 10:
a0 ∈ y, ∃ y0, z. y = y0 + a ' + z ∧ ¬ a ∈ y0 ⊦ dups(y), ∃ y0, a1, z0. a ' + y + a0 ' + z = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Rule argument

spec: list inst: name: firstoc-rew seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y) subst: [x, a] → [y, a] flags: rewrite

Additional information

Used terms: y, a