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Proof step 1 from proof in-union (conclusion)

⊦ a ∈ bag0 ∪ bag1 ↔ a ∈ bag0 ∨ a ∈ bag1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;%union: ⊦ count(a, bag0 ∪ bag1) = count(a, bag0) + count(a, bag1);%union-ass: ⊦ bag0 ∪ bag1 ∪ bag2 = (bag0 ∪ bag1) ∪ bag2;%union-com: ⊦ bag0 ∪ bag1 = bag1 ∪ bag0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2