Metamath Proof Explorer


Theorem nla0001

Description: Extending a linear order to subsets, the empty set is less than itself. Note in Alling, p. 3. (Contributed by RP, 28-Nov-2023)

Ref Expression
Hypothesis nla0001.defsslt < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎𝑆𝑏𝑆 ∧ ∀ 𝑥𝑎𝑦𝑏 𝑥 𝑅 𝑦 ) }
Assertion nla0001 ( 𝜑 → ∅ < ∅ )

Proof

Step Hyp Ref Expression
1 nla0001.defsslt < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎𝑆𝑏𝑆 ∧ ∀ 𝑥𝑎𝑦𝑏 𝑥 𝑅 𝑦 ) }
2 0ex ∅ ∈ V
3 2 a1i ( 𝜑 → ∅ ∈ V )
4 0ss ∅ ⊆ 𝑆
5 4 a1i ( 𝜑 → ∅ ⊆ 𝑆 )
6 1 3 5 nla0002 ( 𝜑 → ∅ < ∅ )