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Proof step 7 from proof ocs-put-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 9:
m ≤ # x0, ∀ m, a, b. m < # x0 → #oc(a, x0[m, b]) = #oc(a, x0) ∨ x0[m] = a ∨ a = b ∨ x0 = [] ⊦ #oc(a0, (a0 ' + x0)[m, b]) = 1 + #oc(a0, x0), a0 ' + x0 = [], (a0 ' + x0).tail[m - 1] = a0, a0 = b, 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);%one-01: ⊦ a ≠ b → #oc(a, b ') = 0;% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (b ' + x)[0, a] = a + x; (s)% ⊦ a + x = a ' + x; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)% ⊦ (a ' + x)[0] = a; (s) Used terms: m, a0 ' + x0