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Proof step 6 from proof shift-03 (previous step)

a ∈ y, ∀ a, y, x0. a ∈ y → (perm(x1 + a ' + x0, y) ↔ perm(x1 + x0, y -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 all left

And obtained the following new goals

Proof step 7:
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)

Rule argument

pos: left 2 terms: a, y -1l a0, x0 abort: F discard: T computed: T

Additional information

Used terms: a, y -1l a0, x0