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Proof step 6 from proof perm-last-rec (previous step)

∀ x, a. perm(x, y0 + a ') ↔ a ∈ x ∧ perm(x -1l a, y0) ⊦ a0 ∈ x ∧ perm(x -1l a0, y0 + a ') ↔ a ∈ x ∧ a0 ∈ x -1l a ∧ perm((x -1l a) -1l a0, y0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ (x -1l a) -1l b = (x -1l b) -1l a
Proof step 8:
∀ x, a. perm(x, y0 + a ') ↔ a ∈ x ∧ perm(x -1l a, y0) ⊦ a0 ∈ x ∧ perm(x -1l a0, y0 + a ') ↔ a ∈ x ∧ a0 ∈ x -1l a ∧ perm((x -1l a0) -1l a, y0)

Rule argument

spec: list-del inst: name: sw-02 seq: ⊦ (x -1l a) -1l b = (x -1l b) -1l a subst: [b, a, x] → [a0, a, x] flags: rewrite

Additional information

Used terms: a0, a, x