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Proof step 5 from proof subset-empty-size (previous step)

bag0 ++ a ⊆ ∅ ⊦ bag0 ++ a = bag0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1))
Proof step 7:
∀ a0. count(a0, bag0 ++ a) ≤ count(a0, ∅) ⊦ bag0 ++ a = bag0

Rule argument

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

Additional information

Used terms: ∅, bag0 ++ a