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Proof step 1 from proof subset-trans (conclusion)

⊦ bag0 ⊆ bag1 ∧ bag1 ⊆ bag2 → bag0 ⊆ bag2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
count(a, bag2) < count(a, bag0), ∀ a. count(a, bag0) ≤ count(a, bag1), ∀ a. count(a, bag1) ≤ count(a, bag2) ⊦ count(a, bag0) = 0

Rule argument

Additional information

Used simplifier rules: count-subset: ⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1));%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat