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Proof step 2 from proof diff-sub-size (previous step)

∀ a. a ∈ s1 → a ∈ s0 ⊦ #(s0 \ s1) = # s0 - # s1

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a. a ∈ ∅ → a ∈ s0 ⊦ #(s0 \ ∅) = # s0 - # ∅
Proof step 5:
∀ s0. (∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ (s ++ a)) = # s0 - #(s ++ a)

Rule argument

indvar: s1

Additional information

Used simplifier rules: set generated by ∅, ++;;