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Proof step 2 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 3:
perm(x, y[n, a]), n < # y ⊦ a ∈ x ∧ perm(x -1l a, y -1l y[n]), y = []
Proof step 10:
a ∈ x ∧ perm(x -1l a, y -1l y[n]), n < # y ⊦ perm(x, y[n, a]), y = []

Rule argument

pos: right 1

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