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Proof step 15 from proof ocs-put-02 (previous step)

m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ #oc((a0 ' + x0)[m], (a0 ' + x0)[m, b]) = #oc((a0 ' + x0)[m], x0) - 1, (a0 ' + x0)[m] = a0, (a0 ' + x0)[m] = b

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 17:
m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ #oc((a0 ' + x0).tail[m - 1], (a0 ' + x0)[m, b]) = #oc((a0 ' + x0).tail[m - 1], x0) - 1, a0 ' + x0 = [], (a0 ' + x0).tail[m - 1] = b, (a0 ' + x0).tail[m - 1] = a0, m = 0

Rule argument

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

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x)[0] = a; (s)% ⊦ (b ' + x)[0, a] = a + x; (s)% ⊦ a + x = a ' + x; (s) Used terms: m, a0 ' + x0