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Proof step 7 from proof subset-empty-size (previous step)

∀ a0. count(a0, bag0 ++ a) ≤ count(a0, ∅) ⊦ bag0 ++ a = bag0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ a0. count(a0, bag0 ++ a) = 0 ⊦ bag0 ++ a = bag0

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ count(a, ∅) = 0; (s)