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Proof step 9 from proof difference (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
count(a, (bag0 -- a0) \ bag) = (count(a, bag) < count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a, bag);0) ⊦ count(a, (bag0 -- a0) \ bag) = (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0)

Rule argument

Additional information

Used simplifier rules: count-del: ⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(a, bag) ≠ 0 ⊃ 1;0);