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Proof step 5 from proof rev-disjoint-01 (previous step)

disj(x, y0), ∀ x. disj(rev(x), y0) ↔ disj(x, y0) ⊦ disj(rev(x), a ' + y0), a ∈ x

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: x

Additional information

Used simplifier rules: rev-in: ⊦ a ∈ rev(x) ↔ a ∈ x;% ⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z); (ws)% ⊦ disj(x, a ') ↔ ¬ a ∈ x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x