Theorem rexabsobidv 45006

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 3156	. . 3 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
3	2	adantr 480	. 2 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑀) → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
4		ralabsod.1	. . 3 ⊢ (𝜑 → Tr 𝑀)
5	4	rexabsod 45004	. 2 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑀) → (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥 ∈ 𝑀 (𝑥 ∈ 𝐴 ∧ 𝜒)))
6	3, 5	bitrd 279	1 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑀) → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝑀 (𝑥 ∈ 𝐴 ∧ 𝜒)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∈ wcel 2111  ∃wrex 3056  Tr wtr 5193

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ral 3048  df-rex 3057  df-v 3438  df-ss 3914  df-uni 4855  df-tr 5194

This theorem is referenced by: (None)