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Proof step 1 from proof sum-of-differences-rem (conclusion)

⊦ sum-of-differences(n, bag -- m) = (m ∈ bag ⊃ sum-of-differences(n, bag) - (n - m);sum-of-differences(n, bag))

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ n, m. sum-of-differences(n, bag0 -- m) = (m ∈ bag0 ⊃ sum-of-differences(n, bag0) - (n - m);sum-of-differences(n, bag0)) ⊦ sum-of-differences(n, (bag0 ++ vtblpnum) -- m) = (m ∈ bag0 ++ vtblpnum ⊃ sum-of-differences(n, bag0 ++ vtblpnum) - (n - m);sum-of-differences(n, bag0 ++ vtblpnum))

Rule argument

indvar: bag

Additional information

Used simplifier rules: sum-of-differences-base: ⊦ sum-of-differences(n, ∅) = 0;%nat-bag generated by ∅, ++;;% ⊦ ∅ -- n = ∅; (s)% ⊦ ¬ n ∈ ∅; (s)