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Proof step 1 from proof union-diff-cancel (conclusion)

⊦ (bag0 ∪ bag1) \ bag1 = bag0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: difference: ⊦ count(a, bag0 \ bag1) = (count(a, bag1) ≤ count(a, bag0) ⊃ count(a, bag0) - count(a, bag1) ;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;%eq-01: ⊦ m + n - m = n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%extension: ⊦ bag0 = bag1 ↔ (∀ a. count(a, bag0) = count(a, bag1)); from specification bag-basic