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Proof step 13 from proof key-cons-ext (previous step)

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])

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: n + 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (nd ' + ndx)[n + 1] = ndx[n]; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list] Used terms: n + 1