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Proof step 15 from proof diff-add-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
count(a, bag1) = 0, count(a0, bag0 ++ a) < count(a0, bag1) ⊦ 0 = count(a0, (bag0 \ bag1) ++ a), count(a0, bag1) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat