Metamath Proof Explorer


Theorem nla0002

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

Ref Expression
Hypotheses nla0001.defslts ⊢ < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎 ⊆ 𝑆 ∧ 𝑏 ⊆ 𝑆 ∧ ∀ 𝑥 ∈ 𝑎 ∀ 𝑦 ∈ 𝑏 𝑥 𝑅 𝑦 ) }
nla0001.set ⊢ ( 𝜑 → 𝐴 ∈ V )
nla0002.sset ⊢ ( 𝜑 → 𝐴 ⊆ 𝑆 )
Assertion nla0002 ( 𝜑 → ∅ < 𝐴 )

Proof

Step Hyp Ref Expression
1 nla0001.defslts ⊢ < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎 ⊆ 𝑆 ∧ 𝑏 ⊆ 𝑆 ∧ ∀ 𝑥 ∈ 𝑎 ∀ 𝑦 ∈ 𝑏 𝑥 𝑅 𝑦 ) }
2 nla0001.set ⊢ ( 𝜑 → 𝐴 ∈ V )
3 nla0002.sset ⊢ ( 𝜑 → 𝐴 ⊆ 𝑆 )
4 0ex ⊢ ∅ ∈ V
5 4 a1i ⊢ ( 𝜑 → ∅ ∈ V )
6 0ss ⊢ ∅ ⊆ 𝑆
7 6 a1i ⊢ ( 𝜑 → ∅ ⊆ 𝑆 )
8 ral0 ⊢ ∀ 𝑥 ∈ ∅ ∀ 𝑦 ∈ 𝐴 𝑥 𝑅 𝑦
9 8 a1i ⊢ ( 𝜑 → ∀ 𝑥 ∈ ∅ ∀ 𝑦 ∈ 𝐴 𝑥 𝑅 𝑦 )
10 7 3 9 3jca ⊢ ( 𝜑 → ( ∅ ⊆ 𝑆 ∧ 𝐴 ⊆ 𝑆 ∧ ∀ 𝑥 ∈ ∅ ∀ 𝑦 ∈ 𝐴 𝑥 𝑅 𝑦 ) )
11 1 rp-brsslt ⊢ ( ∅ < 𝐴 ↔ ( ( ∅ ∈ V ∧ 𝐴 ∈ V ) ∧ ( ∅ ⊆ 𝑆 ∧ 𝐴 ⊆ 𝑆 ∧ ∀ 𝑥 ∈ ∅ ∀ 𝑦 ∈ 𝐴 𝑥 𝑅 𝑦 ) ) )
12 5 2 10 11 syl21anbrc ⊢ ( 𝜑 → ∅ < 𝐴 )