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Proof step 10 from proof ocs-put-01 (previous step)

m ≤ # x0, ∀ m, a, b. m < # x0 → #oc(a, x0[m, b]) = #oc(a, x0) ∨ x0[m] = a ∨ a = b ∨ x0 = [] ⊦ #oc(a0, (a0 ' + x0)[m, b]) = 1 + #oc(a0, x0), x0[m - 1] = a0, a0 = b, m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
m = m0 + 1, m ≤ # x0, ∀ m, a, b. m < # x0 → #oc(a, x0[m, b]) = #oc(a, x0) ∨ x0[m] = a ∨ a = b ∨ x0 = [] ⊦ #oc(a0, (a0 ' + x0)[m, b]) = 1 + #oc(a0, x0), x0[m0] = a0, a0 = b, m = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m1]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1