Theorem rabsspr 32520

Description: Conditions for a restricted class abstraction to be a subset of an unordered pair. (Contributed by Thierry Arnoux, 6-Jul-2025.)

Assertion

Ref	Expression
rabsspr	⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))

Distinct variable groups:   𝑥,𝑋   𝑥,𝑌

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

Proof of Theorem rabsspr

Step	Hyp	Ref	Expression
1		df-rab 3437	. . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)}
2		dfpr2 4646	. . 3 ⊢ {𝑋, 𝑌} = {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)}
3	1, 2	sseq12i 4014	. 2 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌} ↔ {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)})
4		ss2ab 4062	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)} ↔ ∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))
5		impexp 450	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ (𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌))))
6	5	albii 1819	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌))))
7		df-ral 3062	. . 3 ⊢ (∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌))))
8	6, 7	bitr4i 278	. 2 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))
9	3, 4, 8	3bitri 297	1 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∨ wo 848  ∀wal 1538   = wceq 1540   ∈ wcel 2108  {cab 2714  ∀wral 3061  {crab 3436   ⊆ wss 3951  {cpr 4628

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-ral 3062  df-rab 3437  df-v 3482  df-un 3956  df-ss 3968  df-sn 4627  df-pr 4629

This theorem is referenced by:  constrfin  33787