Metamath Proof Explorer


Theorem sneqd

Description: Equality deduction for singletons. (Contributed by NM, 22-Jan-2004)

Ref Expression
Hypothesis sneqd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion sneqd ( 𝜑 → { 𝐴 } = { 𝐵 } )

Proof

Step Hyp Ref Expression
1 sneqd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 sneq ⊢ ( 𝐴 = 𝐵 → { 𝐴 } = { 𝐵 } )
3 1 2 syl ⊢ ( 𝜑 → { 𝐴 } = { 𝐵 } )