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Proof step 2 from proof first (previous step)

⊦ disj(x, a ' + y) ↔ disj(x, y), a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ disj(x, y) ↔ (∀ a. ¬(a ∈ x ∧ a ∈ y))
Proof step 4:
⊦ (∀ a0. ¬(a0 ∈ x ∧ a0 ∈ a ' + y)) ↔ disj(x, y), a ∈ x

Rule argument

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

Additional information

Used terms: a ' + y, x