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Proof step 19 from proof bag-rem-empty (previous step)

count(n, bag) = m + 1, m = 0, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n), count(n, bag) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
count(n, bag) = 1, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic