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Proof step 5 from proof subset-plusminus (previous step)

⊦ (∀ a0. count(a0, bag0) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0)) ↔ bag0 ++ a ⊆ bag1, count(a, bag1) = 0

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) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0)) ↔ (∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1)), count(a, bag1) = 0

Rule argument

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

Additional information

Used terms: bag1, bag0 ++ a