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Proof step 2 from proof max-subseteq (previous step)

bag1.max < bag0.max, bag0 ⊆ bag1 ⊦ bag0 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
bag1.max < bag0.max, bag0 ⊆ bag1 ⊦ bag0 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic