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Proof step 13 from proof difference (previous step)

count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag);0) ⊦ count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) + 1 < count(a0, bag0) ⊃ count(a0, bag0) - count(a0, bag) + 1;0)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 14:
count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag);0), count(a0, bag) + 1 < count(a0, bag0) ⊦ count(a0, (bag0 -- a0) \ bag) = count(a0, bag0) - count(a0, bag) + 1
Proof step 16:
count(a0, (bag0 -- a0) \ bag) = (count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a0, bag);0) ⊦ count(a0, (bag0 -- a0) \ bag) = 0, count(a0, bag) + 1 < count(a0, bag0)

Rule argument

pos: right 1

Proof step 13 from proof difference (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information