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Proof step 5 from proof firstN-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2
Proof step 7:
list2set(x) ∩ s = ∅, list2set(x) ∩ s0 = ∅, a ∈ x, a ∈ firstn(n, x), a ∈ s ∨ a ∈ s0, n < # x ⊦ s ∪ s0 = ∅, #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, x = [], #(s ∪ s0) = 0

Rule argument

seq: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2 subst: [s2, s1, a] → [s0, s, a] rotate: F spec: set-union inst: name: Union

Additional information

Used terms: s0, s, a