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Proof step 2 from proof firstN-01 (previous step)

list2set(x) ∩ s0 = ∅, list2set(x) ∩ s = ∅, n < # x ⊦ #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, list2set(firstn(n, x)) ∩ (s ∪ s0) = ∅, x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 4:
list2set(x) ∩ s0 = ∅, list2set(x) ∩ s = ∅, n < # x ⊦ #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, ∀ a. a ∈ list2set(firstn(n, x)) ∩ (s ∪ s0) ↔ a ∈ ∅, x = []

Rule argument

seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [∅, list2set(firstn(n, x)) ∩ (s ∪ s0)] rotate: F spec: set-basic inst: name: Extension

Additional information

Used terms: ∅, list2set(firstn(n, x)) ∩ (s ∪ s0)