Theorem rabsspr 32984

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 3415	. . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)}
2		dfpr2 4608	. . 3 ⊢ {𝑋, 𝑌} = {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)}
3	1, 2	sseq12i 3964	. 2 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌} ↔ {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)})
4		ss2ab 4012	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)} ↔ ∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))
5		impexp 456	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ (𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌))))
6	5	albii 1852	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌))))
7		df-ral 3079	. . 3 ⊢ (∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌))))
8	6, 7	bitr4i 281	. 2 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))
9	3, 4, 8	3bitri 300	1 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861  ∀wal 1568   = wceq 1570   ∈ wcel 2145  {cab 2740  ∀wral 3078  {crab 3414   ⊆ wss 3902  {cpr 4589

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 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ral 3079  df-rab 3415  df-v 3455  df-un 3907  df-ss 3919  df-sn 4588  df-pr 4590

This theorem is used by:  constrfin  34264