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Proof step 16 from proof subset-plusminus (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 17:
a0 = a, ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a0, bag0) ≤ count(a0, bag1) - 1
Proof step 19:
∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ a0 = a, count(a, bag1) = 0, count(a0, bag0) ≤ count(a0, bag1) - 0

Rule argument

pos: right 2

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