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Proof step 3 from proof bag-remadd-in (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
∀ n. (bag0 -- n) ++ n = (n ∈ bag0 ⊃ bag0;bag0 ++ n) ⊦ (bag0 -- n) ++ m ++ n = (n ∈ bag0 ++ m ⊃ bag0 ++ m;bag0 ++ m ++ n), m = n

Rule argument

pos: right 1

Additional information

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