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.defsslt < ˙ = 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.defsslt < ˙ = 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