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Proof step 5 from proof lastoc (previous step)

a0 ∈ x0, ∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ ∃ b, y, y0. (b = a0 ∨ b ∈ x1) ∧ x0 = y + b ' + y0 ∧ disj(a0 ' + x1, y0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ x ↔ (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z)
Proof step 7:
∃ y, z. x0 = y + a0 ' + z ∧ ¬ a0 ∈ z, ∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ ∃ b, y, y0. (b = a0 ∨ b ∈ x1) ∧ x0 = y + b ' + y0 ∧ disj(a0 ' + x1, y0)

Rule argument

spec: list inst: name: lastoc-rew seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z) subst: [x, a] → [x0, a0] flags: rewrite

Additional information

Used terms: x0, a0