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

∀ a. count(a, bag0) ≤ count(a, bag) ⊦ bag0 ⊆ bag ++ a

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:
∀ a. count(a, bag0) ≤ count(a, bag) ⊦ ∀ a0. count(a0, bag0) ≤ count(a0, bag ++ a)

Rule argument

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

Additional information

Used terms: bag ++ a, bag0