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Proof step 7 from proof cons-02 (previous step)

disj(x, y0), ∀ x, a. disj(x, y0) → (disj(y0, a ' + x) ↔ ¬ a ∈ y0) ⊦ a = a0, a0 ∈ x, disj(y0, x), a ∈ y0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: x, a

Additional information

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