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Proof step 10 from proof key-cons-app (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: ndx1

Additional information

Used terms: ndx1