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Proof step 5 from proof first-01 (previous step)

⊦ (∀ a0. ¬ a0 ∈ y ∨ ¬ a0 ∈ x) ↔ disj(y, x), a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ disj(x, y) ↔ (∀ a. ¬(a ∈ x ∧ a ∈ y))
Proof step 7:
⊦ (∀ a0. ¬ a0 ∈ y ∨ ¬ a0 ∈ x) ↔ (∀ a. ¬(a ∈ y ∧ a ∈ x)), a ∈ x

Rule argument

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

Additional information

Used terms: x, y