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Proof step 7 from proof subset-plusminus (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 8:
∀ a0. count(a0, bag0) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0) ⊦ count(a, bag1) = 0, ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1)
Proof step 15:
∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, ∀ a0. count(a0, bag0) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0)

Rule argument

pos: right 1

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