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Proof step 6 from proof bag-remadd-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ n. (bag0 -- n) ++ n = (n ∈ bag0 ⊃ bag0;bag0 ++ n) ⊦ bag0 -- n = bag0, m = n, n ∈ bag0

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ bag0 ++ n0 ↔ n ∈ bag0 ∨ n = n0; (ws)% ⊦ bag ++ n = bag2 ++ n ↔ bag = bag2; (s)