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Proof step 2 from proof intersection-size (previous step)

# s0 + # s < #(s0 ∩ s) ⊦ #(s0 ∩ s) = 0, s0 ∩ s = ∅

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅, # s0 + #(s1 ++ a) < #(s0 ∩ (s1 ++ a)) ⊦ #(s0 ∩ (s1 ++ a)) = 0, s0 ∩ (s1 ++ a) = ∅

Rule argument

indvar: s

Additional information

Used simplifier rules: int-e: ⊦ s ∩ ∅ = ∅;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%set generated by ∅, ++;;%Size-empty: ⊦ # ∅ = 0; from specification set-basic