Theorem rabsstp 32826

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

Assertion

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

Distinct variable groups:   𝑥,𝑋   𝑥,𝑌   𝑥,𝑍

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

Proof of Theorem rabsstp

Step	Hyp	Ref	Expression
1		df-rab 3417	. . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)}
2		dftp2 4658	. . 3 ⊢ {𝑋, 𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)}
3	1, 2	sseq12i 3968	. 2 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌, 𝑍} ↔ {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)})
4		ss2ab 4016	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} ↔ ∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))
5		impexp 455	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ (𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
6	5	albii 1849	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
7		df-ral 3080	. . 3 ⊢ (∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
8	6, 7	bitr4i 281	. 2 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))
9	3, 4, 8	3bitri 300	1 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌, 𝑍} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   ∨ w3o 1102  ∀wal 1568   = wceq 1570   ∈ wcel 2143  {cab 2741  ∀wral 3079  {crab 3416   ⊆ wss 3906  {ctp 4594

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3or 1104  df-tru 1573  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-ral 3080  df-rab 3417  df-v 3457  df-un 3911  df-ss 3923  df-sn 4591  df-pr 4593  df-tp 4595

This theorem is referenced by: (None)