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Proof step 3 from proof count-del-same (previous step)

⊦ count(a, bag) = 0, count(a, bag -- a) = count(a, bag) - 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
count(a, bag) = m + 1 ⊦ count(a, bag -- a) = m, count(a, bag) = 0

Rule argument

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

Additional information

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