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Proof step 13 from proof difference (previous step)

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

Rule argument

pos: right 1

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2