Metamath Proof Explorer


Theorem invsym2

Description: The inverse relation is symmetric. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses invfval.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
invfval.n ⊢ 𝑁 = ( Inv ‘ 𝐶 )
invfval.c ⊢ ( 𝜑 → 𝐶 ∈ Cat )
invss.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
invss.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
Assertion invsym2 ( 𝜑 → ◡ ( 𝑋 𝑁 𝑌 ) = ( 𝑌 𝑁 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 invss.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
5 invss.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
6 eqid ⊢ ( Hom ‘ 𝐶 ) = ( Hom ‘ 𝐶 )
7 1 2 3 5 4 6 invss ⊢ ( 𝜑 → ( 𝑌 𝑁 𝑋 ) ⊆ ( ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) × ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) ) )
8 relxp ⊢ Rel ( ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) × ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) )
9 relss ⊢ ( ( 𝑌 𝑁 𝑋 ) ⊆ ( ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) × ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) ) → ( Rel ( ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) × ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) ) → Rel ( 𝑌 𝑁 𝑋 ) ) )
10 7 8 9 mpisyl ⊢ ( 𝜑 → Rel ( 𝑌 𝑁 𝑋 ) )
11 relcnv ⊢ Rel ◡ ( 𝑋 𝑁 𝑌 )
12 10 11 jctil ⊢ ( 𝜑 → ( Rel ◡ ( 𝑋 𝑁 𝑌 ) ∧ Rel ( 𝑌 𝑁 𝑋 ) ) )
13 1 2 3 4 5 invsym ⊢ ( 𝜑 → ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ↔ 𝑔 ( 𝑌 𝑁 𝑋 ) 𝑓 ) )
14 vex ⊢ 𝑔 ∈ V
15 vex ⊢ 𝑓 ∈ V
16 14 15 brcnv ⊢ ( 𝑔 ◡ ( 𝑋 𝑁 𝑌 ) 𝑓 ↔ 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 )
17 df-br ⊢ ( 𝑔 ◡ ( 𝑋 𝑁 𝑌 ) 𝑓 ↔ ⟨ 𝑔 , 𝑓 ⟩ ∈ ◡ ( 𝑋 𝑁 𝑌 ) )
18 16 17 bitr3i ⊢ ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ↔ ⟨ 𝑔 , 𝑓 ⟩ ∈ ◡ ( 𝑋 𝑁 𝑌 ) )
19 df-br ⊢ ( 𝑔 ( 𝑌 𝑁 𝑋 ) 𝑓 ↔ ⟨ 𝑔 , 𝑓 ⟩ ∈ ( 𝑌 𝑁 𝑋 ) )
20 13 18 19 3bitr3g ⊢ ( 𝜑 → ( ⟨ 𝑔 , 𝑓 ⟩ ∈ ◡ ( 𝑋 𝑁 𝑌 ) ↔ ⟨ 𝑔 , 𝑓 ⟩ ∈ ( 𝑌 𝑁 𝑋 ) ) )
21 20 eqrelrdv2 ⊢ ( ( ( Rel ◡ ( 𝑋 𝑁 𝑌 ) ∧ Rel ( 𝑌 𝑁 𝑋 ) ) ∧ 𝜑 ) → ◡ ( 𝑋 𝑁 𝑌 ) = ( 𝑌 𝑁 𝑋 ) )
22 12 21 mpancom ⊢ ( 𝜑 → ◡ ( 𝑋 𝑁 𝑌 ) = ( 𝑌 𝑁 𝑋 ) )