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Proof step 14 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, bag) + 1 < count(a0, bag0) ⊦ count(a0, (bag0 -- a0) \ bag) = count(a0, bag0) - count(a0, bag) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
count(a0, (bag0 -- a0) \ bag) = count(a0, bag0) - 1 - count(a0, bag), count(a0, bag) + 1 < count(a0, bag0) ⊦ count(a0, bag0) - 1 - count(a0, bag) = count(a0, bag0) - count(a0, bag) + 1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → n ≠ 0; (s)