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Proof step 9 from proof diff-add-sw (previous step)

count(a0, bag1) ≤ count(a0, bag0 ++ a), count(a, bag1) = 0 ⊦ count(a0, bag0 ++ a) - count(a0, bag1) = count(a0, (bag0 \ bag1) ++ a)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
count(a0, bag0 ++ a) = n0 + count(a0, bag1), count(a0, bag1) ≤ count(a0, bag0 ++ a), count(a, bag1) = 0 ⊦ n0 = count(a0, (bag0 \ bag1) ++ a)

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [count(a0, bag0 ++ a), count(a0, bag1), n] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: count(a0, bag0 ++ a), count(a0, bag1), n