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Proof step 18 from proof subset-plusminus (previous step)

a0 ≠ a, count(a0, bag1) - 0 < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a0, bag0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
count(a0, bag1) < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a0, bag0) = 0, a0 = a

Rule argument

Additional information

Used simplifier rules: sub-base: ⊦ m - 0 = m; from specification nat