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Proof step 1 from proof in-add (conclusion)

⊦ a ∈ bag ++ b ↔ a ∈ bag ∨ a = b

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;%in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%count-other: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag); from specification bag-basic