Theorem bj-sngleq 38057

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 3319	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑦 ∈ 𝐴 𝑥 = {𝑦} ↔ ∃𝑦 ∈ 𝐵 𝑥 = {𝑦}))
2	1	abbidv 2829	. 2 ⊢ (𝐴 = 𝐵 → {𝑥 ∣ ∃𝑦 ∈ 𝐴 𝑥 = {𝑦}} = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = {𝑦}})
3		df-bj-sngl 38056	. 2 ⊢ sngl 𝐴 = {𝑥 ∣ ∃𝑦 ∈ 𝐴 𝑥 = {𝑦}}
4		df-bj-sngl 38056	. 2 ⊢ sngl 𝐵 = {𝑥 ∣ ∃𝑦 ∈ 𝐵 𝑥 = {𝑦}}
5	2, 3, 4	3eqtr4g 2823	1 ⊢ (𝐴 = 𝐵 → sngl 𝐴 = sngl 𝐵)

This proof depends on syntax axioms:   → wi 4   = wceq 1570  {cab 2741  ∃wrex 3090  {csn 4588  sngl bj-csngl 38055

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-rex 3091  df-bj-sngl 38056

This theorem is used by:  bj-tageq  38066