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Proof step 14 from proof ocs-put (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 16:
m ≤ # x0, ∀ m, a. m < # x0 → #oc(a, x0[m, a]) = #oc(a, x0) + 1 ∨ x0[m] = a ∨ x0 = [] ⊦ #oc(a, (a0 ' + x0)[m, a]) = 1 + #oc(a, x0), a0 ' + x0 = [], (a0 ' + x0).tail[m - 1] = a, a = 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);%one: ⊦ #oc(a, a ') = 1;% ⊦ 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)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, a0 ' + x0