Metamath Proof Explorer
Description: If G is an inverse to F , then G is an isomorphism.
(Contributed by Mario Carneiro, 3-Jan-2017)
|
|
Ref |
Expression |
|
Hypotheses |
invfval.b |
⊢ 𝐵 = ( Base ‘ 𝐶 ) |
|
|
invfval.n |
⊢ 𝑁 = ( Inv ‘ 𝐶 ) |
|
|
invfval.c |
⊢ ( 𝜑 → 𝐶 ∈ Cat ) |
|
|
invfval.x |
⊢ ( 𝜑 → 𝑋 ∈ 𝐵 ) |
|
|
invfval.y |
⊢ ( 𝜑 → 𝑌 ∈ 𝐵 ) |
|
|
isoval.n |
⊢ 𝐼 = ( Iso ‘ 𝐶 ) |
|
|
inviso1.1 |
⊢ ( 𝜑 → 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ) |
|
Assertion |
inviso2 |
⊢ ( 𝜑 → 𝐺 ∈ ( 𝑌 𝐼 𝑋 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
invfval.b |
⊢ 𝐵 = ( Base ‘ 𝐶 ) |
2 |
|
invfval.n |
⊢ 𝑁 = ( Inv ‘ 𝐶 ) |
3 |
|
invfval.c |
⊢ ( 𝜑 → 𝐶 ∈ Cat ) |
4 |
|
invfval.x |
⊢ ( 𝜑 → 𝑋 ∈ 𝐵 ) |
5 |
|
invfval.y |
⊢ ( 𝜑 → 𝑌 ∈ 𝐵 ) |
6 |
|
isoval.n |
⊢ 𝐼 = ( Iso ‘ 𝐶 ) |
7 |
|
inviso1.1 |
⊢ ( 𝜑 → 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ) |
8 |
1 2 3 4 5
|
invsym |
⊢ ( 𝜑 → ( 𝐹 ( 𝑋 𝑁 𝑌 ) 𝐺 ↔ 𝐺 ( 𝑌 𝑁 𝑋 ) 𝐹 ) ) |
9 |
7 8
|
mpbid |
⊢ ( 𝜑 → 𝐺 ( 𝑌 𝑁 𝑋 ) 𝐹 ) |
10 |
1 2 3 5 4 6 9
|
inviso1 |
⊢ ( 𝜑 → 𝐺 ∈ ( 𝑌 𝐼 𝑋 ) ) |