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Proof step 6 from proof still-in-rem (previous step)

⊦ count(a, bag0) - (a = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) = 0, a ∈ bag0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
a = b ⊦ count(a, bag0) - 1 = 0, count(b, bag0) = 0, a ∈ bag0
Proof step 12:
a ≠ b ∨ count(b, bag0) = 0 ⊦ count(a, bag0) - 0 = 0, a ∈ bag0

Rule argument

pos: right 1

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