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

⊦ (∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1 ++ a)) ↔ (∀ a. count(a, bag0) ≤ count(a, bag1))

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1 ++ a) ⊦ ∀ a. count(a, bag0) ≤ count(a, bag1)
Proof step 10:
∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1 ++ a)

Rule argument

pos: right 1

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