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Proof step 5 from proof ocs-put-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b]
Proof step 7:
m ≠ 0, (a0 ' + x0)[m] = a0, m ≤ # x0, ∀ m0, b0. x0[m0] ≠ b0 ∧ m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ⊦ a0 = b, #oc(a0, a0 ' + x0[m -1, b]) = #oc(a0, x0)

Rule argument

spec: list-del inst: 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: rewrite

Additional information

Used simplifier rules: oc-n: ⊦ a ≠ b → #oc(a, b ' + x) = #oc(a, x);%get-zero-01: ⊦ (a ' + x)[0] = a; from specification list-del%a: ⊦ (x + y) + z = x + y + z; from specification list%put-zero: ⊦ (b ' + x)[0, a] = a + x; from specification list-del%c: ⊦ a + x = a ' + x; from specification list Used terms: b, m, x0, a0