Metamath Proof Explorer


Theorem prproropf1olem1

Description: Lemma 1 for prproropf1o . (Contributed by AV, 12-Mar-2023)

Ref Expression
Hypotheses prproropf1o.o ⊢ O = R ∩ V × V
prproropf1o.p ⊢ P = p ∈ 𝒫 V | p = 2
Assertion prproropf1olem1 ⊢ R Or V ∧ W ∈ O → 1 st ⁡ W 2 nd ⁡ W ∈ P

Proof

Step Hyp Ref Expression
1 prproropf1o.o ⊢ O = R ∩ V × V
2 prproropf1o.p ⊢ P = p ∈ 𝒫 V | p = 2
3 1 prproropf1olem0 ⊢ W ∈ O ↔ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W
4 simpr2 ⊢ R Or V ∧ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V
5 prelpwi ⊢ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V → 1 st ⁡ W 2 nd ⁡ W ∈ 𝒫 V
6 4 5 syl ⊢ R Or V ∧ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → 1 st ⁡ W 2 nd ⁡ W ∈ 𝒫 V
7 sopo ⊢ R Or V → R Po V
8 7 adantr ⊢ R Or V ∧ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → R Po V
9 simpr3 ⊢ R Or V ∧ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → 1 st ⁡ W R 2 nd ⁡ W
10 po2ne ⊢ R Po V ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → 1 st ⁡ W ≠ 2 nd ⁡ W
11 8 4 9 10 syl3anc ⊢ R Or V ∧ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → 1 st ⁡ W ≠ 2 nd ⁡ W
12 fvex ⊢ 1 st ⁡ W ∈ V
13 fvex ⊢ 2 nd ⁡ W ∈ V
14 hashprg ⊢ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V → 1 st ⁡ W ≠ 2 nd ⁡ W ↔ 1 st ⁡ W 2 nd ⁡ W = 2
15 12 13 14 mp2an ⊢ 1 st ⁡ W ≠ 2 nd ⁡ W ↔ 1 st ⁡ W 2 nd ⁡ W = 2
16 11 15 sylib ⊢ R Or V ∧ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → 1 st ⁡ W 2 nd ⁡ W = 2
17 6 16 jca ⊢ R Or V ∧ W = 1 st ⁡ W 2 nd ⁡ W ∧ 1 st ⁡ W ∈ V ∧ 2 nd ⁡ W ∈ V ∧ 1 st ⁡ W R 2 nd ⁡ W → 1 st ⁡ W 2 nd ⁡ W ∈ 𝒫 V ∧ 1 st ⁡ W 2 nd ⁡ W = 2
18 3 17 sylan2b ⊢ R Or V ∧ W ∈ O → 1 st ⁡ W 2 nd ⁡ W ∈ 𝒫 V ∧ 1 st ⁡ W 2 nd ⁡ W = 2
19 fveqeq2 ⊢ p = 1 st ⁡ W 2 nd ⁡ W → p = 2 ↔ 1 st ⁡ W 2 nd ⁡ W = 2
20 19 2 elrab2 ⊢ 1 st ⁡ W 2 nd ⁡ W ∈ P ↔ 1 st ⁡ W 2 nd ⁡ W ∈ 𝒫 V ∧ 1 st ⁡ W 2 nd ⁡ W = 2
21 18 20 sylibr ⊢ R Or V ∧ W ∈ O → 1 st ⁡ W 2 nd ⁡ W ∈ P