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Proof step 1 from proof union-size (conclusion)

⊦ #(bag0 ∪ bag1) = # bag0 + # bag1

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ bag1. #(bag ∪ bag1) = # bag + # bag1 ⊦ #((bag ++ a) ∪ bag1) = #(bag ++ a) + # bag1

Rule argument

indvar: bag0

Additional information

Used simplifier rules: union-empty: ⊦ ∅ ∪ bag = bag;%union-ass: ⊦ bag0 ∪ bag1 ∪ bag2 = (bag0 ∪ bag1) ∪ bag2;%union-com: ⊦ bag0 ∪ bag1 = bag1 ∪ bag0;%size-empty: ⊦ # ∅ = 0;%bag generated by ∅, ++;;%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2