Metamath Proof Explorer


Theorem ot1st2nd

Description: Reconstruction of a member of a nested Cartesian product in terms of its ordered triple components. (Contributed by BTernaryTau, 9-Oct-2026)

Ref Expression
Assertion ot1st2nd ⊢ A ∈ B × C × D → A = 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A 2 nd ⁡ A

Proof

Step Hyp Ref Expression
1 1st2nd2 ⊢ A ∈ B × C × D → A = 1 st ⁡ A 2 nd ⁡ A
2 xp1st ⊢ A ∈ B × C × D → 1 st ⁡ A ∈ B × C
3 1st2nd2 ⊢ 1 st ⁡ A ∈ B × C → 1 st ⁡ A = 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A
4 2 3 syl ⊢ A ∈ B × C × D → 1 st ⁡ A = 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A
5 4 opeq1d ⊢ A ∈ B × C × D → 1 st ⁡ A 2 nd ⁡ A = 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A 2 nd ⁡ A
6 1 5 eqtrd ⊢ A ∈ B × C × D → A = 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A 2 nd ⁡ A
7 df-ot ⊢ 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A 2 nd ⁡ A = 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A 2 nd ⁡ A
8 6 7 eqtr4di ⊢ A ∈ B × C × D → A = 1 st ⁡ 1 st ⁡ A 2 nd ⁡ 1 st ⁡ A 2 nd ⁡ A