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Proof step 11 from proof diff-add-sw (previous step)

count(a, bag1) = 0, count(a0, bag0 ++ a) = n0 + count(a0, bag1), count(a0, bag1) ≤ count(a0, bag0 ++ a) ⊦ n0 = count(a0, (bag0 \ bag1) ++ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
count(a, bag1) = 0, count(a0, bag0 ++ a) = n0 + count(a0, bag1) ⊦ n0 = count(a0, (bag0 \ bag1) ++ a)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)