Metamath Proof Explorer


Theorem resubid1

Description: Real number version of subid1 , without ax-mulcom . (Contributed by SN, 23-Jan-2024)

Ref Expression
Assertion resubid1 ( 𝐴 ∈ ℝ → ( 𝐴 0 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 readdid2 ( 𝐴 ∈ ℝ → ( 0 + 𝐴 ) = 𝐴 )
2 id ( 𝐴 ∈ ℝ → 𝐴 ∈ ℝ )
3 elre0re ( 𝐴 ∈ ℝ → 0 ∈ ℝ )
4 2 3 2 resubaddd ( 𝐴 ∈ ℝ → ( ( 𝐴 0 ) = 𝐴 ↔ ( 0 + 𝐴 ) = 𝐴 ) )
5 1 4 mpbird ( 𝐴 ∈ ℝ → ( 𝐴 0 ) = 𝐴 )