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Proof step 6 from proof subset-plusplus (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
count(a0, bag1) < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1 ++ a) ⊦ count(a0, bag0) = 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