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Metamath Proof Explorer

Theorem preq2d

Description: Equality deduction for unordered pairs. (Contributed by NM, 19-Oct-2012)

Ref Expression
Hypothesis preq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion preq2d ( 𝜑 → { 𝐶 , 𝐴 } = { 𝐶 , 𝐵 } )

Proof

Step Hyp Ref Expression
1 preq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 preq2 ⊢ ( 𝐴 = 𝐵 → { 𝐶 , 𝐴 } = { 𝐶 , 𝐵 } )
3 1 2 syl ⊢ ( 𝜑 → { 𝐶 , 𝐴 } = { 𝐶 , 𝐵 } )