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Proof step 15 from proof subset-plusminus (previous step)

∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, ∀ a0. count(a0, bag0) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information