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Proof step 1 from proof cons-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
disj(x, y), a ∈ y, disj(a ' + x, y) ⊦

Rule argument

Additional information

Used simplifier rules: first-01: ⊦ ¬ a ∈ x → (disj(a ' + y, x) ↔ disj(y, x));%a: ⊦ (x + y) + z = x + y + z; from specification list