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Proof step 15 from proof subset-plusminus (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 16:
a0 = a, count(a0, bag1) - 1 < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a0, bag0) = 0
Proof step 18:
a0 ≠ a, count(a0, bag1) - 0 < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a0, bag0) = 0

Rule argument

pos: left 1

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