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Theorem ssprss 4828
Description: A pair as subset of a pair. (Contributed by AV, 26-Oct-2020.)
Assertion
Ref Expression
ssprss ((𝐴𝑉𝐵𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))

Proof of Theorem ssprss
StepHypRef Expression
1 prssg 4823 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ {𝐴, 𝐵} ⊆ {𝐶, 𝐷}))
2 elprg 4650 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐴 = 𝐷)))
3 elprg 4650 . . 3 (𝐵𝑊 → (𝐵 ∈ {𝐶, 𝐷} ↔ (𝐵 = 𝐶𝐵 = 𝐷)))
42, 3bi2anan9 635 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
51, 4bitr3d 280 1 ((𝐴𝑉𝐵𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 394  wo 843   = wceq 1539  wcel 2104  wss 3949  {cpr 4631
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-tru 1542  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-v 3474  df-un 3954  df-in 3956  df-ss 3966  df-sn 4630  df-pr 4632
This theorem is referenced by:  ssprsseq  4829
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