Theorem rmoeqd 3405

Description: Equality deduction for restricted at-most-one quantifier. (Contributed by Alexander van der Vekens, 17-Jun-2017.)

Hypothesis

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

Assertion

Ref	Expression
rmoeqd	⊢ (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 rmoeqd

Step	Hyp	Ref	Expression
1		rmoeq1 3403	. 2 ⊢ (𝐴 = 𝐵 → (∃*𝑥 ∈ 𝐴 𝜑 ↔ ∃*𝑥 ∈ 𝐵 𝜑))
2		rmoeqd.1	. . 3 ⊢ (𝐴 = 𝐵 → (𝜑 ↔ 𝜓))
3	2	rmobidv 3387	. 2 ⊢ (𝐴 = 𝐵 → (∃*𝑥 ∈ 𝐵 𝜑 ↔ ∃*𝑥 ∈ 𝐵 𝜓))
4	1, 3	bitrd 282	1 ⊢ (𝐴 = 𝐵 → (∃*𝑥 ∈ 𝐴 𝜑 ↔ ∃*𝑥 ∈ 𝐵 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570  ∃*wrmo 3371

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 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-mo 2570  df-cleq 2758  df-rmo 3372

This theorem is used by: (None)