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Proof step 15 from proof subset-plusplus (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
count(a0, bag1) < count(a0, bag0), ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ a0 = a, count(a0, bag0) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%count-other: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag); from specification bag-basic