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Proof step 5 from proof intersection-size (previous step)

# s0 + #(s1 ++ a) < #(s0 ∩ (s1 ++ a)), ∀ s0. ¬ # s0 + # s1 < #(s0 ∩ s1) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ s → s ++ a = s
Proof step 7:
# s0 + #(s1 ++ a) < #(s0 ∩ (s1 ++ a)), ∀ s0. ¬ # s0 + # s1 < #(s0 ∩ s1) ⊦ a ∈ s1
Proof step 17:
# s0 + # s1 < #(s0 ∩ s1), a ∈ s1, ∀ s0. ¬ # s0 + # s1 < #(s0 ∩ s1) ⊦

Rule argument

spec: set-basic name: eq seq: ⊦ a ∈ s → s ++ a = s subst: [a, s] → [a, s1] flags: <all paths>

Additional information

Used terms: a, s1