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Proof step 19 from proof subset-plusminus (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
count(a0, bag1) < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ a0 = a, count(a, bag1) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m - 0 = m; (s)