Metamath Proof Explorer


Theorem qseq1

Description: Equality theorem for quotient set. (Contributed by NM, 23-Jul-1995)

Ref Expression
Assertion qseq1 ( 𝐴 = 𝐵 → ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) )

Proof

Step Hyp Ref Expression
1 rexeq ( 𝐴 = 𝐵 → ( ∃ 𝑥𝐴 𝑦 = [ 𝑥 ] 𝐶 ↔ ∃ 𝑥𝐵 𝑦 = [ 𝑥 ] 𝐶 ) )
2 1 abbidv ( 𝐴 = 𝐵 → { 𝑦 ∣ ∃ 𝑥𝐴 𝑦 = [ 𝑥 ] 𝐶 } = { 𝑦 ∣ ∃ 𝑥𝐵 𝑦 = [ 𝑥 ] 𝐶 } )
3 df-qs ( 𝐴 / 𝐶 ) = { 𝑦 ∣ ∃ 𝑥𝐴 𝑦 = [ 𝑥 ] 𝐶 }
4 df-qs ( 𝐵 / 𝐶 ) = { 𝑦 ∣ ∃ 𝑥𝐵 𝑦 = [ 𝑥 ] 𝐶 }
5 2 3 4 3eqtr4g ( 𝐴 = 𝐵 → ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) )