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Proof step 4 from proof intersection-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a ∈ s → s ++ a = s
Proof step 6:
¬ a ∈ s1, # s0 + #(s1 ++ a) < #(s0 ∩ (s1 ++ a)), ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ #(s0 ∩ (s1 ++ a)) = 0, s0 ∩ (s1 ++ a) = ∅
Proof step 16:
a ∈ s1, # s0 + # s1 < #(s0 ∩ s1), ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ #(s0 ∩ s1) = 0, s0 ∩ s1 = ∅

Rule argument

spec: set-basic inst: name: eq seq: ⊦ a ∈ s → s ++ a = s subst: [a, s] → [a, s1] flags: rewrite

Additional information

Used terms: a, s1