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Proof step 14 from proof firstN-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 16:
∀ a. a ∈ list2set(x) ∩ s0 ↔ a ∈ ∅, list2set(x) ∩ s = ∅, n < # x, a ∈ firstn(n, x), a ∈ x, a ∈ s0 ⊦ s0 = ∅, #(s ∪ s0) = 0, x = [], #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, s ∪ s0 = ∅, # s0 = 0

Rule argument

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

Additional information

Used terms: ∅, list2set(x) ∩ s0