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Proof step 10 from proof lastoc (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ disj(x, y) ↔ (∀ a. ¬(a ∈ x ∧ a ∈ y))
Proof step 12:
∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ a0 ∈ z, ∀ a. ¬(a ∈ x1 ∧ a ∈ z), ∃ b, y1, y0. (b = a0 ∨ b ∈ x1) ∧ y + a0 ' + z = y1 + b ' + y0 ∧ disj(a0 ' + x1, y0)

Rule argument

spec: inst: name: disjoint seq: ⊦ disj(x, y) ↔ (∀ a. ¬(a ∈ x ∧ a ∈ y)) subst: [y, x] → [z, x1] flags: rewrite, current

Additional information

Used terms: z, x1