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Proof step 7 from proof sum-of-differences-rem (previous step)

m ∈ bag0, ∀ 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) - (n - m), vtblpnum = m

User applied rule all left

And obtained the following new goals

Proof step 8:
sum-of-differences(n, bag0 -- m) = (m ∈ bag0 ⊃ sum-of-differences(n, bag0) - (n - m);sum-of-differences(n, bag0)), m ∈ bag0 ⊦ (n - vtblpnum) + sum-of-differences(n, bag0 -- m) = (n - vtblpnum) + sum-of-differences(n, bag0) - (n - m), vtblpnum = m

Rule argument

pos: left 2 terms: n, m abort: F discard: T computed: F

Additional information

Used terms: n, m