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Proof step 12 from proof key-cons-ext (previous step)

∀ n. ¬ # ndx0 < n → key-cons((nd ' + ndx0)[n].key, (nd ' + ndx0)[n]) ⊦ key-cons(ndx0), ∀ 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 13:
n < # ndx0, ∀ n. ¬ # ndx0 < n → key-cons((nd ' + ndx0)[n].key, (nd ' + ndx0)[n]) ⊦ ndx0 = [], key-cons(ndx0), key-cons(ndx0[n].key, ndx0[n])

Rule argument

Additional information

Used simplifier rules: ⊦ # ndx = 0 ↔ ndx = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic