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Proof step 16 from proof bag-rm-add-eq (previous step)

a ∈ bag ⊦ 1 + (count(a, bag) - (count(a, bag) ≠ 0 ⊃ 1;0)) = count(a, bag)

User applied rule case distinction

And obtained the following new goals

Proof step 17:
count(a, bag) = 0, a ∈ bag ⊦ 1 + (count(a, bag) - 0) = count(a, bag)

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic