Theorem rexabsobidv 45742

Description: Formula-building lemma for proving absoluteness results. (Contributed by Eric Schmidt, 19-Oct-2025.)

Hypotheses

Ref	Expression
ralabsod.1	⊢ (𝜑 → Tr 𝑀)
ralabsobidv.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
rexabsobidv	⊢ ((𝜑 ∧ 𝐴 ∈ 𝑀) → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝑀 (𝑥 ∈ 𝐴 ∧ 𝜒)))

Distinct variable groups:   𝑥,𝑀   𝑥,𝐴   𝜑,𝑥

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

Proof of Theorem rexabsobidv

Step	Hyp	Ref	Expression
1		ralabsobidv.2	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2	1	rexbidv 3191	. . 3 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
3	2	adantr 486	. 2 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑀) → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
4		ralabsod.1	. . 3 ⊢ (𝜑 → Tr 𝑀)
5	4	rexabsod 45740	. 2 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑀) → (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥 ∈ 𝑀 (𝑥 ∈ 𝐴 ∧ 𝜒)))
6	3, 5	bitrd 282	1 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑀) → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝑀 (𝑥 ∈ 𝐴 ∧ 𝜒)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∈ wcel 2146  ∃wrex 3091  Tr wtr 5220

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-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rex 3092  df-v 3459  df-ss 3923  df-uni 4875  df-tr 5221

This theorem is used by: (None)