Metamath Proof Explorer


Theorem mat1rhmelval

Description: The value of the ring homomorphism F . (Contributed by AV, 22-Dec-2019)

Ref Expression
Hypotheses mat1rhmval.k 𝐾 = ( Base ‘ 𝑅 )
mat1rhmval.a 𝐴 = ( { 𝐸 } Mat 𝑅 )
mat1rhmval.b 𝐵 = ( Base ‘ 𝐴 )
mat1rhmval.o 𝑂 = ⟨ 𝐸 , 𝐸
mat1rhmval.f 𝐹 = ( 𝑥𝐾 ↦ { ⟨ 𝑂 , 𝑥 ⟩ } )
Assertion mat1rhmelval ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → ( 𝐸 ( 𝐹𝑋 ) 𝐸 ) = 𝑋 )

Proof

Step Hyp Ref Expression
1 mat1rhmval.k 𝐾 = ( Base ‘ 𝑅 )
2 mat1rhmval.a 𝐴 = ( { 𝐸 } Mat 𝑅 )
3 mat1rhmval.b 𝐵 = ( Base ‘ 𝐴 )
4 mat1rhmval.o 𝑂 = ⟨ 𝐸 , 𝐸
5 mat1rhmval.f 𝐹 = ( 𝑥𝐾 ↦ { ⟨ 𝑂 , 𝑥 ⟩ } )
6 df-ov ( 𝐸 ( 𝐹𝑋 ) 𝐸 ) = ( ( 𝐹𝑋 ) ‘ ⟨ 𝐸 , 𝐸 ⟩ )
7 1 2 3 4 5 mat1rhmval ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → ( 𝐹𝑋 ) = { ⟨ 𝑂 , 𝑋 ⟩ } )
8 7 fveq1d ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → ( ( 𝐹𝑋 ) ‘ ⟨ 𝐸 , 𝐸 ⟩ ) = ( { ⟨ 𝑂 , 𝑋 ⟩ } ‘ ⟨ 𝐸 , 𝐸 ⟩ ) )
9 4 eqcomi 𝐸 , 𝐸 ⟩ = 𝑂
10 9 fveq2i ( { ⟨ 𝑂 , 𝑋 ⟩ } ‘ ⟨ 𝐸 , 𝐸 ⟩ ) = ( { ⟨ 𝑂 , 𝑋 ⟩ } ‘ 𝑂 )
11 opex 𝐸 , 𝐸 ⟩ ∈ V
12 4 11 eqeltri 𝑂 ∈ V
13 simp3 ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → 𝑋𝐾 )
14 fvsng ( ( 𝑂 ∈ V ∧ 𝑋𝐾 ) → ( { ⟨ 𝑂 , 𝑋 ⟩ } ‘ 𝑂 ) = 𝑋 )
15 12 13 14 sylancr ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → ( { ⟨ 𝑂 , 𝑋 ⟩ } ‘ 𝑂 ) = 𝑋 )
16 10 15 eqtrid ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → ( { ⟨ 𝑂 , 𝑋 ⟩ } ‘ ⟨ 𝐸 , 𝐸 ⟩ ) = 𝑋 )
17 8 16 eqtrd ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → ( ( 𝐹𝑋 ) ‘ ⟨ 𝐸 , 𝐸 ⟩ ) = 𝑋 )
18 6 17 eqtrid ( ( 𝑅 ∈ Ring ∧ 𝐸𝑉𝑋𝐾 ) → ( 𝐸 ( 𝐹𝑋 ) 𝐸 ) = 𝑋 )