Theorem rabsstp 32572

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 3390	. . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)}
2		dftp2 4635	. . 3 ⊢ {𝑋, 𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)}
3	1, 2	sseq12i 3952	. 2 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌, 𝑍} ↔ {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)})
4		ss2ab 4001	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} ↔ ∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))
5		impexp 450	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ (𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
6	5	albii 1821	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
7		df-ral 3052	. . 3 ⊢ (∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
8	6, 7	bitr4i 278	. 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 297	1 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌, 𝑍} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∨ w3o 1086  ∀wal 1540   = wceq 1542   ∈ wcel 2114  {cab 2714  ∀wral 3051  {crab 3389   ⊆ wss 3889  {ctp 4571

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3or 1088  df-tru 1545  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-nfc 2885  df-ral 3052  df-rab 3390  df-v 3431  df-un 3894  df-ss 3906  df-sn 4568  df-pr 4570  df-tp 4572

This theorem is referenced by: (None)