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Proof step 4 from proof rev-disjoint-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
disj(x, y0), ∀ x. disj(rev(x), y0) ↔ disj(x, y0) ⊦ a ∈ x, disj(rev(x), a ' + y0)

Rule argument

Additional information

Used simplifier rules: sym: ⊦ disj(x, y) → disj(y, x); from specification list-dup%sym: ⊦ disj(x, y) → disj(y, x); (forward) from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list%one-01: ⊦ disj(x, a ') ↔ ¬ a ∈ x; from specification list-dup%app-02: ⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z); from specification list-dup