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Proof step 6 from proof ocs-put-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b]
Proof step 8:
(a0 ' + x0)[m] = a0, m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ #oc(a0, a0 ' + x0[m -1, b]) = #oc(a0, x0), a0 = b, m = 0

Rule argument

spec: list-del name: put-rec-02 seq: ⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b] subst: [b, n, x, a] → [b, m, x0, a0] 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)[0] = a; (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 Used terms: b, m, x0, a0