Metamath Proof Explorer


Theorem 1stinr

Description: The first component of the value of a right injection is 1o . (Contributed by AV, 27-Jun-2022)

Ref Expression
Assertion 1stinr ( 𝑋𝑉 → ( 1st ‘ ( inr ‘ 𝑋 ) ) = 1o )

Proof

Step Hyp Ref Expression
1 df-inr inr = ( 𝑥 ∈ V ↦ ⟨ 1o , 𝑥 ⟩ )
2 opeq2 ( 𝑥 = 𝑋 → ⟨ 1o , 𝑥 ⟩ = ⟨ 1o , 𝑋 ⟩ )
3 elex ( 𝑋𝑉𝑋 ∈ V )
4 opex ⟨ 1o , 𝑋 ⟩ ∈ V
5 4 a1i ( 𝑋𝑉 → ⟨ 1o , 𝑋 ⟩ ∈ V )
6 1 2 3 5 fvmptd3 ( 𝑋𝑉 → ( inr ‘ 𝑋 ) = ⟨ 1o , 𝑋 ⟩ )
7 6 fveq2d ( 𝑋𝑉 → ( 1st ‘ ( inr ‘ 𝑋 ) ) = ( 1st ‘ ⟨ 1o , 𝑋 ⟩ ) )
8 1oex 1o ∈ V
9 op1stg ( ( 1o ∈ V ∧ 𝑋𝑉 ) → ( 1st ‘ ⟨ 1o , 𝑋 ⟩ ) = 1o )
10 8 9 mpan ( 𝑋𝑉 → ( 1st ‘ ⟨ 1o , 𝑋 ⟩ ) = 1o )
11 7 10 eqtrd ( 𝑋𝑉 → ( 1st ‘ ( inr ‘ 𝑋 ) ) = 1o )