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Proof step 7 from proof intersection-size (previous step)

# s0 + # s1 + 1 < #(s0 ∩ (s1 ++ a)), ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ #(s0 ∩ (s1 ++ a)) = 0, s0 ∩ (s1 ++ a) = ∅, a ∈ s1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ∈ s → s ∩ (s0 ++ a) = (s ∩ s0) ++ a
Proof step 9:
¬ a ∈ s0, # s0 + # s1 + 1 < #(s0 ∩ (s1 ++ a)), ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ #(s0 ∩ (s1 ++ a)) = 0, s0 ∩ (s1 ++ a) = ∅, a ∈ s1
Proof step 11:
a ∈ s0, # s0 + # s1 + 1 < #((s0 ∩ s1) ++ a), ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ #((s0 ∩ s1) ++ a) = 0, (s0 ∩ s1) ++ a = ∅, a ∈ s1

Rule argument

spec: set-union inst: name: intersection-shift seq: ⊦ a ∈ s → s ∩ (s0 ++ a) = (s ∩ s0) ++ a subst: [a, s0, s] → [a, s1, s0] flags: rewrite, current

Additional information

Used terms: a, s1, s0