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Proof step 2 from proof bag-rem-empty (previous step)

⊦ bag -- n = ∅ ↔ bag = ∅ ++ n, bag = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
bag -- n = ∅ ⊦ bag = ∅, bag = ∅ ++ n

Rule argument

Additional information

Used simplifier rules: ⊦ (bag0 ++ n) -- n0 = (n = n0 ⊃ bag0;(bag0 -- n0) ++ n); (s)