Metamath Proof Explorer


Theorem oteqimp

Description: The components of an ordered triple. (Contributed by Alexander van der Vekens, 2-Mar-2018)

Ref Expression
Assertion oteqimp ⊢ T = A B C → A ∈ X ∧ B ∈ Y ∧ C ∈ Z → 1 st ⁡ 1 st ⁡ T = A ∧ 2 nd ⁡ 1 st ⁡ T = B ∧ 2 nd ⁡ T = C

Proof

Step Hyp Ref Expression
1 ot1stg ⊢ A ∈ X ∧ B ∈ Y ∧ C ∈ Z → 1 st ⁡ 1 st ⁡ A B C = A
2 ot2ndg ⊢ A ∈ X ∧ B ∈ Y ∧ C ∈ Z → 2 nd ⁡ 1 st ⁡ A B C = B
3 ot3rdg ⊢ C ∈ Z → 2 nd ⁡ A B C = C
4 3 3ad2ant3 ⊢ A ∈ X ∧ B ∈ Y ∧ C ∈ Z → 2 nd ⁡ A B C = C
5 1 2 4 3jca ⊢ A ∈ X ∧ B ∈ Y ∧ C ∈ Z → 1 st ⁡ 1 st ⁡ A B C = A ∧ 2 nd ⁡ 1 st ⁡ A B C = B ∧ 2 nd ⁡ A B C = C
6 2fveq3 ⊢ T = A B C → 1 st ⁡ 1 st ⁡ T = 1 st ⁡ 1 st ⁡ A B C
7 6 eqeq1d ⊢ T = A B C → 1 st ⁡ 1 st ⁡ T = A ↔ 1 st ⁡ 1 st ⁡ A B C = A
8 2fveq3 ⊢ T = A B C → 2 nd ⁡ 1 st ⁡ T = 2 nd ⁡ 1 st ⁡ A B C
9 8 eqeq1d ⊢ T = A B C → 2 nd ⁡ 1 st ⁡ T = B ↔ 2 nd ⁡ 1 st ⁡ A B C = B
10 fveqeq2 ⊢ T = A B C → 2 nd ⁡ T = C ↔ 2 nd ⁡ A B C = C
11 7 9 10 3anbi123d ⊢ T = A B C → 1 st ⁡ 1 st ⁡ T = A ∧ 2 nd ⁡ 1 st ⁡ T = B ∧ 2 nd ⁡ T = C ↔ 1 st ⁡ 1 st ⁡ A B C = A ∧ 2 nd ⁡ 1 st ⁡ A B C = B ∧ 2 nd ⁡ A B C = C
12 5 11 imbitrrid ⊢ T = A B C → A ∈ X ∧ B ∈ Y ∧ C ∈ Z → 1 st ⁡ 1 st ⁡ T = A ∧ 2 nd ⁡ 1 st ⁡ T = B ∧ 2 nd ⁡ T = C