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Proof step 2 from proof papp-03 (previous step)

perm(x, x0), perm(x0, x) ⊦ perm(x + y, x0 + y0) ↔ perm(y, y0)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
perm(x + y, x0 + y0), perm(x, x0), perm(x0, x) ⊦ perm(y, y0)
Proof step 7:
perm(y, y0), perm(x, x0), perm(x0, x) ⊦ perm(x + y, x0 + y0)

Rule argument

pos: right 1

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