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Proof step 1 from proof diff-size (conclusion)

bag1 ⊆ bag0 ⊦ #(bag0 \ bag1) = # bag0 - # bag1

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ bag0. bag ⊆ bag0 → #(bag0 \ bag) = # bag0 - # bag, bag ++ a ⊆ bag0 ⊦ #(bag0 \ (bag ++ a)) = # bag0 - #(bag ++ a)

Rule argument

indvar: bag1

Additional information

Used simplifier rules: diff-empty: ⊦ bag0 \ ∅ = bag0;%size-empty: ⊦ # ∅ = 0;%bag generated by ∅, ++;;%sub-base: ⊦ m - 0 = m; from specification nat