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Proof step 9 from proof still-in-rem (previous step)

count(b, bag0) = m + 1 ⊦ count(b, bag0) = 0, m = 0, b ∈ bag0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
count(b, bag0) = m + 1 ⊦ m = 0, b ∈ bag0

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% ⊦ m + 1 ≠ 0; (s)