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Proof step 20 from proof ocs-put-02 (previous step)

#oc(x0[m1], x0) = m + 1 ∧ #oc(x0[m1], x0) ≠ 0, m1 < # x0, ∀ m0, b0. x0[m0] ≠ b0 ∧ m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ⊦ #oc(x0[m1], x0[m1, b]) = m, x0[m1] = a0, x0[m1] = b, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
#oc(x0[m1], x0) = m + 1, m1 < # x0, x0[m1] ∈ x0, ∀ m0, b0. x0[m0] ≠ b0 ∧ m0 < # x0 → #oc(x0[m0], x0[m0, b0]) = #oc(x0[m0], x0) - 1 ⊦ #oc(x0[m1], x0[m1, b]) = m, x0[m1] = a0, x0[m1] = b, x0 = []

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;