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Proof step 14 from proof less-same (previous step)

m + 1 < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ less#(a, (a0 ' + x0[m, x0[m + 1]])[2 + m, x0[m]]) = (a0 < a ⊃ 1;0) + less#(a, x0), x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a]
Proof step 16:
m + 1 < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ a0 ' + x0[m, x0[m + 1]] = [], 2 + m = 0, less#(a, (a0 ' + x0[m, x0[m + 1]]).head ' + (a0 ' + x0[m, x0[m + 1]]).tail[(2 + m)-1, x0[m]]) = (a0 < a ⊃ 1;0) + less#(a, x0), x0 = []

Rule argument

spec: list-del name: put-rec seq: ⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a] subst: [a, n, x] → [x0[m], 2 + m, a0 ' + x0[m, x0[m + 1]]] flags: <all paths>

Additional information

Used simplifier rules: n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0[m], 2 + m, a0 ' + x0[m, x0[m + 1]]