Theorem reueqd 3125

Description: Equality deduction for restricted uniqueness quantifier. (Contributed by NM, 5-Apr-2004.)

Hypothesis

Ref	Expression
raleqd.1	⊢ (𝐴 = 𝐵 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
reueqd	⊢ (𝐴 = 𝐵 → (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥 ∈ 𝐵 𝜓))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)

Proof of Theorem reueqd

Step	Hyp	Ref	Expression
1		reueq1 3117	. 2 ⊢ (𝐴 = 𝐵 → (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥 ∈ 𝐵 𝜑))
2		raleqd.1	. . 3 ⊢ (𝐴 = 𝐵 → (𝜑 ↔ 𝜓))
3	2	reubidv 3103	. 2 ⊢ (𝐴 = 𝐵 → (∃!𝑥 ∈ 𝐵 𝜑 ↔ ∃!𝑥 ∈ 𝐵 𝜓))
4	1, 3	bitrd 267	1 ⊢ (𝐴 = 𝐵 → (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥 ∈ 𝐵 𝜓))

Syntax hints:   → wi 4   ↔ wb 195   = wceq 1475  ∃!wreu 2898

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-eu 2462  df-cleq 2603  df-clel 2606  df-nfc 2740  df-reu 2903

This theorem is referenced by:  aceq1  8823  isfrgr  41430