Metamath Proof Explorer


Theorem tpeq3d

Description: Equality theorem for unordered triples. (Contributed by NM, 22-Jun-2014)

Ref Expression
Hypothesis tpeq1d.1 ( 𝜑𝐴 = 𝐵 )
Assertion tpeq3d ( 𝜑 → { 𝐶 , 𝐷 , 𝐴 } = { 𝐶 , 𝐷 , 𝐵 } )

Proof

Step Hyp Ref Expression
1 tpeq1d.1 ( 𝜑𝐴 = 𝐵 )
2 tpeq3 ( 𝐴 = 𝐵 → { 𝐶 , 𝐷 , 𝐴 } = { 𝐶 , 𝐷 , 𝐵 } )
3 1 2 syl ( 𝜑 → { 𝐶 , 𝐷 , 𝐴 } = { 𝐶 , 𝐷 , 𝐵 } )