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Proof step 11 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)) ⊦ (n - vtblpnum) + sum-of-differences(n, bag0 -- m) = (n - vtblpnum) + sum-of-differences(n, bag0), vtblpnum = m, m ∈ bag0 ++ vtblpnum

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ 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 -- m) = sum-of-differences(n, bag0), vtblpnum = m, m ∈ bag0

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ bag0 ++ n0 ↔ n ∈ bag0 ∨ n = n0; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)