Metamath Proof Explorer


Theorem isoeq145d

Description: Equality deduction for isometries. (Contributed by RP, 14-Jan-2025)

Ref Expression
Hypotheses isoeq145.1 ( 𝜑𝐹 = 𝐺 )
isoeq145.4 ( 𝜑𝐴 = 𝐶 )
isoeq145.5 ( 𝜑𝐵 = 𝐷 )
Assertion isoeq145d ( 𝜑 → ( 𝐹 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐷 ) ) )

Proof

Step Hyp Ref Expression
1 isoeq145.1 ( 𝜑𝐹 = 𝐺 )
2 isoeq145.4 ( 𝜑𝐴 = 𝐶 )
3 isoeq145.5 ( 𝜑𝐵 = 𝐷 )
4 isoeq1 ( 𝐹 = 𝐺 → ( 𝐹 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) )
5 1 4 syl ( 𝜑 → ( 𝐹 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ) )
6 isoeq4 ( 𝐴 = 𝐶 → ( 𝐺 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐵 ) ) )
7 2 6 syl ( 𝜑 → ( 𝐺 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐵 ) ) )
8 isoeq5 ( 𝐵 = 𝐷 → ( 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐷 ) ) )
9 3 8 syl ( 𝜑 → ( 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐷 ) ) )
10 5 7 9 3bitrd ( 𝜑 → ( 𝐹 Isom 𝑅 , 𝑆 ( 𝐴 , 𝐵 ) ↔ 𝐺 Isom 𝑅 , 𝑆 ( 𝐶 , 𝐷 ) ) )