Metamath Proof Explorer


Theorem nla0003

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

Ref Expression
Hypotheses nla0001.defsslt < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎𝑆𝑏𝑆 ∧ ∀ 𝑥𝑎𝑦𝑏 𝑥 𝑅 𝑦 ) }
nla0001.set ( 𝜑𝐴 ∈ V )
nla0002.sset ( 𝜑𝐴𝑆 )
Assertion nla0003 ( 𝜑𝐴 < ∅ )

Proof

Step Hyp Ref Expression
1 nla0001.defsslt < = { ⟨ 𝑎 , 𝑏 ⟩ ∣ ( 𝑎𝑆𝑏𝑆 ∧ ∀ 𝑥𝑎𝑦𝑏 𝑥 𝑅 𝑦 ) }
2 nla0001.set ( 𝜑𝐴 ∈ V )
3 nla0002.sset ( 𝜑𝐴𝑆 )
4 0ex ∅ ∈ V
5 4 a1i ( 𝜑 → ∅ ∈ V )
6 0ss ∅ ⊆ 𝑆
7 6 a1i ( 𝜑 → ∅ ⊆ 𝑆 )
8 ral0 𝑦 ∈ ∅ ∀ 𝑥𝐴 𝑥 𝑅 𝑦
9 ralcom ( ∀ 𝑦 ∈ ∅ ∀ 𝑥𝐴 𝑥 𝑅 𝑦 ↔ ∀ 𝑥𝐴𝑦 ∈ ∅ 𝑥 𝑅 𝑦 )
10 8 9 mpbi 𝑥𝐴𝑦 ∈ ∅ 𝑥 𝑅 𝑦
11 10 a1i ( 𝜑 → ∀ 𝑥𝐴𝑦 ∈ ∅ 𝑥 𝑅 𝑦 )
12 3 7 11 3jca ( 𝜑 → ( 𝐴𝑆 ∧ ∅ ⊆ 𝑆 ∧ ∀ 𝑥𝐴𝑦 ∈ ∅ 𝑥 𝑅 𝑦 ) )
13 1 rp-brsslt ( 𝐴 < ∅ ↔ ( ( 𝐴 ∈ V ∧ ∅ ∈ V ) ∧ ( 𝐴𝑆 ∧ ∅ ⊆ 𝑆 ∧ ∀ 𝑥𝐴𝑦 ∈ ∅ 𝑥 𝑅 𝑦 ) ) )
14 2 5 12 13 syl21anbrc ( 𝜑𝐴 < ∅ )