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Proof step 2 from proof sum-of-differences-rem (previous step)

∀ 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ 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, (vtblpnum = m ⊃ bag0;(bag0 -- m) ++ vtblpnum)) = (m ∈ bag0 ++ vtblpnum ⊃ (n - vtblpnum) + sum-of-differences(n, bag0) - (n - m);(n - vtblpnum) + sum-of-differences(n, bag0))

Rule argument

Additional information

Used simplifier rules: sum-of-differences-rec: ⊦ sum-of-differences(n, bag ++ m) = (n - m) + sum-of-differences(n, bag);% ⊦ (bag0 ++ n) -- n0 = (n = n0 ⊃ bag0;(bag0 -- n0) ++ n); (s)