Metamath Proof Explorer


Theorem fuco2eld2

Description: Equivalence of product functor. (Contributed by Zhi Wang, 29-Sep-2025)

Ref Expression
Hypotheses fuco2eld.w ⊢ φ → W = S × R
fuco2eld2.u ⊢ φ → U ∈ W
fuco2eld2.s ⊢ Rel ⁡ S
fuco2eld2.r ⊢ Rel ⁡ R
Assertion fuco2eld2 ⊢ φ → U = 1 st ⁡ 1 st ⁡ U 2 nd ⁡ 1 st ⁡ U 1 st ⁡ 2 nd ⁡ U 2 nd ⁡ 2 nd ⁡ U

Proof

Step Hyp Ref Expression
1 fuco2eld.w ⊢ φ → W = S × R
2 fuco2eld2.u ⊢ φ → U ∈ W
3 fuco2eld2.s ⊢ Rel ⁡ S
4 fuco2eld2.r ⊢ Rel ⁡ R
5 2 1 eleqtrd ⊢ φ → U ∈ S × R
6 1st2nd2 ⊢ U ∈ S × R → U = 1 st ⁡ U 2 nd ⁡ U
7 5 6 syl ⊢ φ → U = 1 st ⁡ U 2 nd ⁡ U
8 df-rel ⊢ Rel ⁡ S ↔ S ⊆ V × V
9 3 8 mpbi ⊢ S ⊆ V × V
10 xp1st ⊢ U ∈ S × R → 1 st ⁡ U ∈ S
11 9 10 sselid ⊢ U ∈ S × R → 1 st ⁡ U ∈ V × V
12 1st2nd2 ⊢ 1 st ⁡ U ∈ V × V → 1 st ⁡ U = 1 st ⁡ 1 st ⁡ U 2 nd ⁡ 1 st ⁡ U
13 5 11 12 3syl ⊢ φ → 1 st ⁡ U = 1 st ⁡ 1 st ⁡ U 2 nd ⁡ 1 st ⁡ U
14 df-rel ⊢ Rel ⁡ R ↔ R ⊆ V × V
15 4 14 mpbi ⊢ R ⊆ V × V
16 xp2nd ⊢ U ∈ S × R → 2 nd ⁡ U ∈ R
17 15 16 sselid ⊢ U ∈ S × R → 2 nd ⁡ U ∈ V × V
18 1st2nd2 ⊢ 2 nd ⁡ U ∈ V × V → 2 nd ⁡ U = 1 st ⁡ 2 nd ⁡ U 2 nd ⁡ 2 nd ⁡ U
19 5 17 18 3syl ⊢ φ → 2 nd ⁡ U = 1 st ⁡ 2 nd ⁡ U 2 nd ⁡ 2 nd ⁡ U
20 13 19 opeq12d ⊢ φ → 1 st ⁡ U 2 nd ⁡ U = 1 st ⁡ 1 st ⁡ U 2 nd ⁡ 1 st ⁡ U 1 st ⁡ 2 nd ⁡ U 2 nd ⁡ 2 nd ⁡ U
21 7 20 eqtrd ⊢ φ → U = 1 st ⁡ 1 st ⁡ U 2 nd ⁡ 1 st ⁡ U 1 st ⁡ 2 nd ⁡ U 2 nd ⁡ 2 nd ⁡ U