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Proof step 1 from proof rec (conclusion)

⊦ a ≠ b → (perm(a ' + x, b ' + y) ↔ a ∈ y ∧ b ∈ x ∧ perm(x -1l b, y -1l a))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-01: ⊦ perm(a ' + x, y) ↔ ¬ ¬(a ∈ y ∧ perm(x, y -1l a));%del-03: ⊦ perm(x, a ' + y) ↔ ¬ ¬(a ∈ x ∧ perm(x -1l a, y));%del: ⊦ perm(x, a ' + y) ↔ a ∈ x ∧ perm(x -1l a, y);%del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); from specification list-del%a: ⊦ (x + y) + z = x + y + z; from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list