Step |
Hyp |
Ref |
Expression |
1 |
|
fgraphopab |
⊢ ( 𝐹 : 𝐴 ⟶ 𝐵 → 𝐹 = { 〈 𝑎 , 𝑏 〉 ∣ ( ( 𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵 ) ∧ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) } ) |
2 |
|
vex |
⊢ 𝑎 ∈ V |
3 |
|
vex |
⊢ 𝑏 ∈ V |
4 |
2 3
|
op1std |
⊢ ( 𝑥 = 〈 𝑎 , 𝑏 〉 → ( 1st ‘ 𝑥 ) = 𝑎 ) |
5 |
4
|
fveq2d |
⊢ ( 𝑥 = 〈 𝑎 , 𝑏 〉 → ( 𝐹 ‘ ( 1st ‘ 𝑥 ) ) = ( 𝐹 ‘ 𝑎 ) ) |
6 |
2 3
|
op2ndd |
⊢ ( 𝑥 = 〈 𝑎 , 𝑏 〉 → ( 2nd ‘ 𝑥 ) = 𝑏 ) |
7 |
5 6
|
eqeq12d |
⊢ ( 𝑥 = 〈 𝑎 , 𝑏 〉 → ( ( 𝐹 ‘ ( 1st ‘ 𝑥 ) ) = ( 2nd ‘ 𝑥 ) ↔ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) ) |
8 |
7
|
rabxp |
⊢ { 𝑥 ∈ ( 𝐴 × 𝐵 ) ∣ ( 𝐹 ‘ ( 1st ‘ 𝑥 ) ) = ( 2nd ‘ 𝑥 ) } = { 〈 𝑎 , 𝑏 〉 ∣ ( 𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵 ∧ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) } |
9 |
|
df-3an |
⊢ ( ( 𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵 ∧ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) ↔ ( ( 𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵 ) ∧ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) ) |
10 |
9
|
opabbii |
⊢ { 〈 𝑎 , 𝑏 〉 ∣ ( 𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵 ∧ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) } = { 〈 𝑎 , 𝑏 〉 ∣ ( ( 𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵 ) ∧ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) } |
11 |
8 10
|
eqtri |
⊢ { 𝑥 ∈ ( 𝐴 × 𝐵 ) ∣ ( 𝐹 ‘ ( 1st ‘ 𝑥 ) ) = ( 2nd ‘ 𝑥 ) } = { 〈 𝑎 , 𝑏 〉 ∣ ( ( 𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵 ) ∧ ( 𝐹 ‘ 𝑎 ) = 𝑏 ) } |
12 |
1 11
|
eqtr4di |
⊢ ( 𝐹 : 𝐴 ⟶ 𝐵 → 𝐹 = { 𝑥 ∈ ( 𝐴 × 𝐵 ) ∣ ( 𝐹 ‘ ( 1st ‘ 𝑥 ) ) = ( 2nd ‘ 𝑥 ) } ) |