Metamath Proof Explorer
Description: Equality deduction for functions. (Contributed by Glauco Siliprandi, 11-Dec-2019)
|
|
Ref |
Expression |
|
Hypotheses |
feq1dd.eq |
⊢ ( 𝜑 → 𝐹 = 𝐺 ) |
|
|
feq1dd.f |
⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 ) |
|
Assertion |
feq1dd |
⊢ ( 𝜑 → 𝐺 : 𝐴 ⟶ 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
feq1dd.eq |
⊢ ( 𝜑 → 𝐹 = 𝐺 ) |
2 |
|
feq1dd.f |
⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 ) |
3 |
1
|
feq1d |
⊢ ( 𝜑 → ( 𝐹 : 𝐴 ⟶ 𝐵 ↔ 𝐺 : 𝐴 ⟶ 𝐵 ) ) |
4 |
2 3
|
mpbid |
⊢ ( 𝜑 → 𝐺 : 𝐴 ⟶ 𝐵 ) |