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Proof step 4 from proof trans (previous step)

∀ a. #oc(a, z) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, z), ∀ a. #oc(a, y) ≤ #oc(a, x), ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ x ⊆m z ∧ z ⊆m x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a. #oc(a, z) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, z), ∀ a. #oc(a, y) ≤ #oc(a, x), ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ #oc(a, x) ≤ #oc(a, z) ∧ #oc(a0, z) ≤ #oc(a0, x)

Rule argument

Additional information

Used simplifier rules: msubset: ⊦ x ⊆m y ↔ (∀ a. #oc(a, x) ≤ #oc(a, y));