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Proof step 14 from proof bag-rm-add-eq (previous step)

a ∈ bag ⊦ (a0 = a ⊃ 1;0) + (count(a0, bag) - (a0 = a ∧ count(a, bag) ≠ 0 ⊃ 1;0)) = count(a0, bag)

User applied rule case distinction

And obtained the following new goals

Proof step 15:
a0 = a, a ∈ bag ⊦ 1 + (count(a0, bag) - (a0 = a ∧ count(a, bag) ≠ 0 ⊃ 1;0)) = count(a0, bag)

Rule argument

pos: right 1

Additional information

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