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Proof step 14 from proof key-cons-ext (previous step)

key-cons(ndx0), ∀ n. ¬ # ndx0 < n → key-cons((nd ' + ndx0)[n].key, (nd ' + ndx0)[n]), ∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n]) ⊦ key-cons(nd.key, nd) ∧ key-cons(ndx0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
key-cons(ndx0), ∀ n. n < # ndx0 → key-cons(ndx0[n].key, ndx0[n]), ∀ n. ¬ # ndx0 < n → key-cons((nd ' + ndx0)[n].key, (nd ' + ndx0)[n]) ⊦ key-cons(nd.key, nd)

Proof step 14 from proof key-cons-ext (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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