Theorem rabsstp 32597

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 3393	. . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)}
2		dftp2 4630	. . 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 3999	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)} ↔ ∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))
5		impexp 451	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ (𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
6	5	albii 1826	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
7		df-ral 3055	. . 3 ⊢ (∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍))))
8	6, 7	bitr4i 279	. 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 298	1 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌, 𝑍} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))

Syntax hints:   → wi 4   ↔ wb 207   ∧ wa 396   ∨ w3o 1091  ∀wal 1545   = wceq 1547   ∈ wcel 2119  {cab 2718  ∀wral 3054  {crab 3392   ⊆ wss 3890  {ctp 4566

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-10 2152  ax-11 2168  ax-12 2189  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3or 1093  df-tru 1550  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-nfc 2889  df-ral 3055  df-rab 3393  df-v 3434  df-un 3895  df-ss 3907  df-sn 4563  df-pr 4565  df-tp 4567

This theorem is referenced by: (None)