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Proof step 2 from proof subset-emp (previous step)

bag ⊆ ∅ ⊦ bag = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bag = ∅ ↔ ¬(∃ a. a ∈ bag)
Proof step 4:
bag ⊆ ∅ ⊦ ∀ a. ¬ a ∈ bag

Rule argument

spec: <toplevel> name: some seq: ⊦ bag = ∅ ↔ ¬(∃ a. a ∈ bag) subst: [bag] → [bag] flags: <all paths>

Additional information

Used terms: bag