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Proof step 5 from proof bag-remadd-in (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
∀ n. (bag0 -- n) ++ n = (n ∈ bag0 ⊃ bag0;bag0 ++ n) ⊦ (bag0 -- n) ++ m ++ n = bag0 ++ m ++ n, m = n, n ∈ bag0 ++ m

Rule argument

pos: right 1

Additional information

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