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Proof step 3 from proof subset-elem (previous step)

a ∈ bag0, bag0 ⊆ bag1 ⊦ a ∈ bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1))
Proof step 5:
a ∈ bag0, ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ a ∈ bag1

Rule argument

spec: <toplevel> name: count-subset seq: ⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1)) subst: [bag1, bag0] → [bag1, bag0] flags: <all paths>

Additional information

Used terms: bag1, bag0