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Proof step 5 from proof first (previous step)

⊦ (∀ a0. ¬(a0 ∈ x ∧ a0 ∈ a ' + y)) ↔ disj(x, y), a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)