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Proof step 5 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) = (m ∈ bag0 ++ vtblpnum ⊃ (n - vtblpnum) + sum-of-differences(n, bag0) - (n - m);(n - vtblpnum) + sum-of-differences(n, bag0)), vtblpnum = m

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
m ∈ bag0 ++ vtblpnum, ∀ 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
Proof step 11:
∀ 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

Rule argument

pos: right 1

Proof step 5 from proof sum-of-differences-rem (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information