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Proof step 3 from proof max-subseteq (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 4:
bag1.max < (bag ++ m).max, bag ++ m ⊆ bag1, ∀ bag1. bag1.max < bag.max ∧ bag ⊆ bag1 → bag = ∅ ⊦ bag ++ m = ∅

Rule argument

indvar: bag0

Additional information

Used simplifier rules: nat-bag generated by ∅, ++;;