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Proof step 15 from proof key-cons-ext (previous step)

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)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: 0

Additional information

Used simplifier rules: ⊦ ¬ m < 0; (s)% ⊦ (nd ' + ndx)[0] = nd; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list] Used terms: 0