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Proof step 22 from proof lastoc (previous step)

disj(y0, x1), disj(x1, y0), b ∈ x1, a ∈ x1, a0 ∈ y0, ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)), a ∈ y ∨ a = b ⊦ a ∈ y0, b ∈ y0, a0 ∈ x1, ∃ b0, y1, y2. (b0 = a0 ∨ b0 ∈ x1) ∧ y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(a0 ' + x1, y2)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ a ∈ x ↔ (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z)
Proof step 24:
disj(x1, y0), disj(y0, x1), a ∈ x1, b ∈ x1, ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)), ∃ y, z. y0 = y + a0 ' + z ∧ ¬ a0 ∈ z, a ∈ y ∨ a = b ⊦ a0 ∈ x1, b ∈ y0, a ∈ y0, ∃ b0, y1, y2. (b0 = a0 ∨ b0 ∈ x1) ∧ y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(a0 ' + x1, y2)

Rule argument

spec: list name: lastoc-rew seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ z) subst: [x, a] → [y0, a0] flags: <all paths>

Additional information

Used terms: y0, a0