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Proof step 26 from proof bag-add-rem-same (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
∀ n. (∀ n0. count(n0, (bag0 -- n) ++ n) = count(n0, bag0)) ↔ count(n, bag0) ≠ 0 ⊦ (n0 = n ⊃ 1;0) + count(n0, bag0 -- n) = count(n0, bag0), count(n, bag0) = 0, n = m, n0 = m

Rule argument

Additional information

Used simplifier rules: ⊦ n ≠ n0 → count(n, bag0 ++ n0) = count(n, bag0); (s)