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Proof step 11 from proof bag-rem-empty (previous step)

∀ n0. count(n0, bag -- n) = count(n0, ∅) ⊦ count(n0, bag) = 0, count(n1, bag) = count(n1, ∅ ++ n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n), count(n0, bag) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ count(n, ∅) = 0; (s)