Theorem bj-sngleq

Description: Substitution property for sngl . (Contributed by BJ, 6-Oct-2018)

	Ref	Expression
Assertion	bj-sngleq	⊢ ( 𝐴 = 𝐵 → sngl 𝐴 = sngl 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		rexeq	⊢ ( 𝐴 = 𝐵 → ( ∃ 𝑦 ∈ 𝐴 𝑥 = { 𝑦 } ↔ ∃ 𝑦 ∈ 𝐵 𝑥 = { 𝑦 } ) )
2	1	abbidv	⊢ ( 𝐴 = 𝐵 → { 𝑥 ∣ ∃ 𝑦 ∈ 𝐴 𝑥 = { 𝑦 } } = { 𝑥 ∣ ∃ 𝑦 ∈ 𝐵 𝑥 = { 𝑦 } } )
3		df-bj-sngl	⊢ sngl 𝐴 = { 𝑥 ∣ ∃ 𝑦 ∈ 𝐴 𝑥 = { 𝑦 } }
4		df-bj-sngl	⊢ sngl 𝐵 = { 𝑥 ∣ ∃ 𝑦 ∈ 𝐵 𝑥 = { 𝑦 } }
5	2 3 4	3eqtr4g	⊢ ( 𝐴 = 𝐵 → sngl 𝐴 = sngl 𝐵 )