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Proof step 2 from proof bag-remadd-in (previous step)

∀ n. (bag0 -- n) ++ n = (n ∈ bag0 ⊃ bag0;bag0 ++ n) ⊦ ((bag0 ++ m) -- n) ++ n = (n ∈ bag0 ++ m ⊃ bag0 ++ m;bag0 ++ m ++ n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. (bag0 -- n) ++ n = (n ∈ bag0 ⊃ bag0;bag0 ++ n) ⊦ (m = n ⊃ bag0;(bag0 -- n) ++ m) ++ n = (n ∈ bag0 ++ m ⊃ bag0 ++ m;bag0 ++ m ++ n)

Rule argument

Additional information

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