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Proof step 4 from proof subset-one (previous step)

count(a, bag) = 0, ∅ ++ a ⊆ bag ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1))
Proof step 6:
count(a, bag) = 0, ∀ a0. count(a0, ∅ ++ a) ≤ count(a0, bag) ⊦

Rule argument

spec: inst: name: count-subset seq: ⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1)) subst: [bag1, bag0] → [bag, ∅ ++ a] flags: rewrite, current

Additional information

Used terms: bag, ∅ ++ a