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Proof step 9 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 ⊦ 1 + #oc(a0, x0[m - 1, b]) = #oc(a0, x0), a0 = b, m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
(a0 ' + x0)[m] = a0, m = m1 + 1, m ≤ # x0, ∀ m0, b0. m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ∨ x0[m0] = b0 ⊦ 1 + #oc(a0, x0[m1, b]) = #oc(a0, x0), 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