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Mathbox for Thierry Arnoux |
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Mirrors > Home > MPE Home > Th. List > Mathboxes > rabsspr | Structured version Visualization version GIF version |
Description: Conditions for a restricted class abstraction to be a subset of an unordered pair. (Contributed by Thierry Arnoux, 6-Jul-2025.) |
Ref | Expression |
---|---|
rabsspr | ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-rab 3434 | . . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} | |
2 | dfpr2 4651 | . . 3 ⊢ {𝑋, 𝑌} = {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)} | |
3 | 1, 2 | sseq12i 4026 | . 2 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋, 𝑌} ↔ {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)}) |
4 | ss2ab 4072 | . 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 1816 | . . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → (𝑥 = 𝑋 ∨ 𝑥 = 𝑌)))) |
7 | df-ral 3060 | . . 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))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 206 ∧ wa 395 ∨ wo 847 ∀wal 1535 = wceq 1537 ∈ wcel 2106 {cab 2712 ∀wral 3059 {crab 3433 ⊆ wss 3963 {cpr 4633 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-8 2108 ax-9 2116 ax-10 2139 ax-11 2155 ax-12 2175 ax-ext 2706 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-tru 1540 df-ex 1777 df-nf 1781 df-sb 2063 df-clab 2713 df-cleq 2727 df-clel 2814 df-nfc 2890 df-ral 3060 df-rab 3434 df-v 3480 df-un 3968 df-ss 3980 df-sn 4632 df-pr 4634 |
This theorem is referenced by: constrfin 33751 |
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