Metamath Proof Explorer


Theorem predeq123

Description: Equality theorem for the predecessor class. (Contributed by Scott Fenton, 13-Jun-2018)

Ref Expression
Assertion predeq123 ( ( 𝑅 = 𝑆𝐴 = 𝐵𝑋 = 𝑌 ) → Pred ( 𝑅 , 𝐴 , 𝑋 ) = Pred ( 𝑆 , 𝐵 , 𝑌 ) )

Proof

Step Hyp Ref Expression
1 simp2 ( ( 𝑅 = 𝑆𝐴 = 𝐵𝑋 = 𝑌 ) → 𝐴 = 𝐵 )
2 cnveq ( 𝑅 = 𝑆 𝑅 = 𝑆 )
3 2 3ad2ant1 ( ( 𝑅 = 𝑆𝐴 = 𝐵𝑋 = 𝑌 ) → 𝑅 = 𝑆 )
4 sneq ( 𝑋 = 𝑌 → { 𝑋 } = { 𝑌 } )
5 4 3ad2ant3 ( ( 𝑅 = 𝑆𝐴 = 𝐵𝑋 = 𝑌 ) → { 𝑋 } = { 𝑌 } )
6 3 5 imaeq12d ( ( 𝑅 = 𝑆𝐴 = 𝐵𝑋 = 𝑌 ) → ( 𝑅 “ { 𝑋 } ) = ( 𝑆 “ { 𝑌 } ) )
7 1 6 ineq12d ( ( 𝑅 = 𝑆𝐴 = 𝐵𝑋 = 𝑌 ) → ( 𝐴 ∩ ( 𝑅 “ { 𝑋 } ) ) = ( 𝐵 ∩ ( 𝑆 “ { 𝑌 } ) ) )
8 df-pred Pred ( 𝑅 , 𝐴 , 𝑋 ) = ( 𝐴 ∩ ( 𝑅 “ { 𝑋 } ) )
9 df-pred Pred ( 𝑆 , 𝐵 , 𝑌 ) = ( 𝐵 ∩ ( 𝑆 “ { 𝑌 } ) )
10 7 8 9 3eqtr4g ( ( 𝑅 = 𝑆𝐴 = 𝐵𝑋 = 𝑌 ) → Pred ( 𝑅 , 𝐴 , 𝑋 ) = Pred ( 𝑆 , 𝐵 , 𝑌 ) )