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Proof step 8 from proof diff-add-sw (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
count(a0, bag1) ≤ count(a0, bag0 ++ a), count(a, bag1) = 0 ⊦ count(a0, bag0 ++ a) - count(a0, bag1) = count(a0, (bag0 \ bag1) ++ a)
Proof step 15:
¬ count(a0, bag1) ≤ count(a0, bag0 ++ a), count(a, bag1) = 0 ⊦ 0 = count(a0, (bag0 \ bag1) ++ a)

Rule argument

pos: right 1

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