Metamath Proof Explorer


Theorem predeq1

Description: Equality theorem for the predecessor class. (Contributed by Scott Fenton, 2-Feb-2011)

Ref Expression
Assertion predeq1 ( 𝑅 = 𝑆 → Pred ( 𝑅 , 𝐴 , 𝑋 ) = Pred ( 𝑆 , 𝐴 , 𝑋 ) )

Proof

Step Hyp Ref Expression
1 eqid ⊢ 𝐴 = 𝐴
2 eqid ⊢ 𝑋 = 𝑋
3 predeq123 ⊢ ( ( 𝑅 = 𝑆 ∧ 𝐴 = 𝐴 ∧ 𝑋 = 𝑋 ) → Pred ( 𝑅 , 𝐴 , 𝑋 ) = Pred ( 𝑆 , 𝐴 , 𝑋 ) )
4 1 2 3 mp3an23 ⊢ ( 𝑅 = 𝑆 → Pred ( 𝑅 , 𝐴 , 𝑋 ) = Pred ( 𝑆 , 𝐴 , 𝑋 ) )