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Proof step 11 from proof subset-antisym (previous step)

count(a, bag0) ≤ count(a, bag1), ∀ a. count(a, bag0) ≤ count(a, bag1), ∀ a. count(a, bag1) ≤ count(a, bag0) ⊦ count(a, bag0) = count(a, bag1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
count(a, bag0) < count(a, bag1), ∀ a. count(a, bag0) ≤ count(a, bag1), ∀ a. count(a, bag1) ≤ count(a, bag0) ⊦ count(a, bag0) = count(a, bag1)

Rule argument

Additional information

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