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Proof step 19 from proof difference (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) = n0 + count(a0, bag), count(a0, (bag0 -- a0) \ bag) = n0, count(a0, bag) < count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag0) ≤ count(a0, bag) + 1, count(a0, bag0) ≤ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag) ≤ count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) ⊦ count(a0, (bag0 -- a0) \ bag) = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag), n]

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0), count(a0, bag), n