Metamath Proof Explorer


Theorem xnegeq

Description: Equality of two extended numbers with -e in front of them. (Contributed by FL, 26-Dec-2011) (Proof shortened by Mario Carneiro, 20-Aug-2015)

Ref Expression
Assertion xnegeq ( 𝐴 = 𝐵 → -𝑒 𝐴 = -𝑒 𝐵 )

Proof

Step Hyp Ref Expression
1 eqeq1 ( 𝐴 = 𝐵 → ( 𝐴 = +∞ ↔ 𝐵 = +∞ ) )
2 eqeq1 ( 𝐴 = 𝐵 → ( 𝐴 = -∞ ↔ 𝐵 = -∞ ) )
3 negeq ( 𝐴 = 𝐵 → - 𝐴 = - 𝐵 )
4 2 3 ifbieq2d ( 𝐴 = 𝐵 → if ( 𝐴 = -∞ , +∞ , - 𝐴 ) = if ( 𝐵 = -∞ , +∞ , - 𝐵 ) )
5 1 4 ifbieq2d ( 𝐴 = 𝐵 → if ( 𝐴 = +∞ , -∞ , if ( 𝐴 = -∞ , +∞ , - 𝐴 ) ) = if ( 𝐵 = +∞ , -∞ , if ( 𝐵 = -∞ , +∞ , - 𝐵 ) ) )
6 df-xneg -𝑒 𝐴 = if ( 𝐴 = +∞ , -∞ , if ( 𝐴 = -∞ , +∞ , - 𝐴 ) )
7 df-xneg -𝑒 𝐵 = if ( 𝐵 = +∞ , -∞ , if ( 𝐵 = -∞ , +∞ , - 𝐵 ) )
8 5 6 7 3eqtr4g ( 𝐴 = 𝐵 → -𝑒 𝐴 = -𝑒 𝐵 )