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Proof step 12 from proof bag-add-rem-same (previous step)

(n = m ⊃ 1;0) + count(n, bag0) = 0, ∀ n. (∀ n0. count(n0, (bag0 -- n) ++ n) = count(n0, bag0)) ↔ count(n, bag0) ≠ 0, ∀ n0. count(n0, (m = n ⊃ bag0;(bag0 -- n) ++ m) ++ n) = count(n0, bag0 ++ m) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
count(n, bag0) = 0, ∀ n. (∀ n0. count(n0, (bag0 -- n) ++ n) = count(n0, bag0)) ↔ count(n, bag0) ≠ 0, ∀ n0. count(n0, bag0 ++ m ++ n) = count(n0, bag0 ++ m) ⊦ n = m

Rule argument

Additional information

Used simplifier rules: ⊦ count(n, bag0) = 0 → bag0 -- n = bag0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)