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Proof step 4 from proof sum-of-differences-in (previous step)

sum-of-differences(n, bag0 ++ vtblpnum) < n - m, m ∈ bag0 ++ vtblpnum, ∀ n, m. ¬(sum-of-differences(n, bag0) < n - m ∧ m ∈ bag0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(n - vtblpnum) + sum-of-differences(n, bag0) < n - m, m ∈ bag0 ++ vtblpnum, ∀ n0, m0. ¬ sum-of-differences(n0, bag0) < n0 - m0 ∨ ¬ m0 ∈ 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);