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Proof step 4 from proof less-same (previous step)

n < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ less#(a, (a0 ' + x0)[n, x0[n]][n + 1, (a0 ' + x0)[n]]) = (a0 < a ⊃ 1;0) + less#(a, x0), x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 6:
n < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ less#(a, (a0 ' + x0)[n, x0[n]][n + 1, (a0 ' + x0)[n]]) = (a0 < a ⊃ 1;0) + less#(a, x0), x0 = [], a0 ' + x0 ≠ [] ∧ n ≠ 0
Proof step 11:
n < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ a0 ' + x0 = [], less#(a, (a0 ' + x0)[n, x0[n]][n + 1, (a0 ' + x0).tail[n - 1]]) = (a0 < a ⊃ 1;0) + less#(a, x0), n = 0, x0 = []

Rule argument

spec: list-del name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1] subst: [n, x] → [n, a0 ' + x0] flags: <all paths>

Additional information

Used terms: n, a0 ' + x0