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Proof step 6 from proof ocs-put-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

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

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: oc-n: ⊦ a ≠ b → #oc(a, b ' + x) = #oc(a, x);%e-01: ⊦ a ' + x ≠ []; from specification list%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%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%get-zero-01: ⊦ (a ' + x)[0] = a; from specification list-del Used terms: m, a0 ' + x0