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Proof step 7 from proof shift-03 (previous step)

a ∈ y, a ∈ y -1l a0 → (perm(x1 + a ' + x0, y -1l a0) ↔ perm(x1 + x0, (y -1l a0) -1l a)) ⊦ a0 ∈ y ∧ perm(x1 + a ' + x0, y -1l a0) ↔ a0 ∈ y -1l a ∧ perm(x1 + x0, (y -1l a) -1l a0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ (x -1l a) -1l b = (x -1l b) -1l a
Proof step 9:
a ∈ y, a ∈ y -1l a0 → (perm(x1 + a ' + x0, y -1l a0) ↔ perm(x1 + x0, (y -1l a0) -1l a)) ⊦ a0 ∈ y ∧ perm(x1 + a ' + x0, y -1l a0) ↔ a0 ∈ y -1l a ∧ perm(x1 + x0, (y -1l a0) -1l a)

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, y] flags: rewrite

Additional information

Used terms: a0, a, y