Theorem bj-sngleq 36304

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

Assertion

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

Proof of Theorem bj-sngleq

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		rexeq 3313	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑦 ∈ 𝐴 𝑥 = {𝑦} ↔ ∃𝑦 ∈ 𝐵 𝑥 = {𝑦}))
2	1	abbidv 2793	. 2 ⊢ (𝐴 = 𝐵 → {𝑥 ∣ ∃𝑦 ∈ 𝐴 𝑥 = {𝑦}} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = {𝑦}})
3		df-bj-sngl 36303	. 2 ⊢ sngl 𝐴 = {𝑥 ∣ ∃𝑦 ∈ 𝐴 𝑥 = {𝑦}}
4		df-bj-sngl 36303	. 2 ⊢ sngl 𝐵 = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = {𝑦}}
5	2, 3, 4	3eqtr4g 2789	1 ⊢ (𝐴 = 𝐵 → sngl 𝐴 = sngl 𝐵)

Syntax hints:   → wi 4   = wceq 1533  {cab 2701  ∃wrex 3062  {csn 4620  sngl bj-csngl 36302

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716  df-rex 3063  df-bj-sngl 36303

This theorem is referenced by:  bj-tageq  36313