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Proof step 9 from proof key-cons-app (previous step)

key-cons(nd.key, nd), key-cons(ndx), ∀ ndx0. key-cons(ndx + ndx0) ↔ key-cons(ndx0) ⊦ key-cons(ndx + ndx1) ↔ key-cons(ndx1), key-cons(ndx + ndx1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information