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Proof step 5 from proof l (previous step)

y ⊆m x → # y ≤ # x, ∀ a. #oc(a, y) ≤ #oc(a, x), ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ # x = # y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# y < # x, ∀ a. #oc(a, y) ≤ #oc(a, x), ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ # x = # y, x = []

Rule argument

Additional information

Used simplifier rules: msubset: ⊦ x ⊆m y ↔ (∀ a. #oc(a, x) ≤ #oc(a, y));% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)