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.defslts ⊢ < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎 ⊆ 𝑆 ∧ 𝑏 ⊆ 𝑆 ∧ ∀ 𝑥 ∈ 𝑎 ∀ 𝑦 ∈ 𝑏 𝑥 𝑅 𝑦 ) }
Assertion nla0001 ( 𝜑 → ∅ < ∅ )

Proof

Step Hyp Ref Expression
1 nla0001.defslts ⊢ < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎 ⊆ 𝑆 ∧ 𝑏 ⊆ 𝑆 ∧ ∀ 𝑥 ∈ 𝑎 ∀ 𝑦 ∈ 𝑏 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 → ∅ < ∅ )