Theorem rexabsobidv 46105

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 3190	. . 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 46103	. 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 3090  Tr wtr 5215

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-v 3456  df-ss 3901  df-uni 4871  df-tr 5216

This theorem is used by: (None)