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Proof step 1 from proof subset-smaller (conclusion)

⊦ bag0 ++ a ⊆ bag1 → (¬ bag0 ⊆ bag1 ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bag0 ⊆ bag ⊦ bag0 ⊆ bag ++ a

Rule argument

Additional information

Used simplifier rules: subset-plus: ⊦ ¬ ¬ bag0 ++ a ⊆ bag1 ↔ (∃ bag2. bag1 = bag2 ++ a ∧ bag0 ⊆ bag2);