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Proof step 3 from proof perm-put (previous step)

n < # y ⊦ perm(x, y[n, a]) ↔ a ∈ x ∧ perm(x -1l a, y -1l y[n]), y = []

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 4:
n < # y, perm(x, y[n, a]) ⊦ y = [], a ∈ x ∧ perm(x -1l a, y -1l y[n])
Proof step 11:
n < # y, perm(x -1l a, y -1l y[n]), a ∈ x ⊦ y = [], perm(x, y[n, a])

Rule argument

pos: right 1

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