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Proof step 18 from proof bag-rem-empty (previous step)

count(n, bag) - 1 = 0, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n, bag) = 0, count(n1, bag) = count(n1, ∅ ++ n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
count(n, bag) = m + 1, m = 0, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n), count(n, bag) = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [count(n, bag), m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: count(n, bag), m