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Proof step 14 from proof ocs-put-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1]
Proof step 16:
a0 ' + x0 ≠ [] ∧ m ≠ 0, m ≤ # x0, ∀ m0, b0. x0[m0] ≠ b0 ∧ m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ⊦ (a0 ' + x0).rest[m - 1] = b, (a0 ' + x0).rest[m - 1] = a0, #oc((a0 ' + x0).rest[m - 1], (a0 ' + x0)[m, b]) = #oc((a0 ' + x0).rest[m - 1], x0) - 1

Rule argument

spec: list-del inst: name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1] subst: [n, x] → [m, a0 ' + x0] flags: rewrite

Additional information

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