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