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Proof step 11 from proof key-cons-app (
previous step
)
key-cons(nd.key, nd), ∀ ndx1. key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1) ⊦ key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1), key-cons(ndx)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 12
:
key-cons(nd.key, nd), key-cons(ndx + ndx1), ∀ ndx0. ¬ key-cons(ndx + ndx0) ⊦ key-cons(ndx)
Rule argument
Additional information