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Theorem ssprss 4849
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 4844 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ {𝐴, 𝐵} ⊆ {𝐶, 𝐷}))
2 elprg 4670 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐴 = 𝐷)))
3 elprg 4670 . . 3 (𝐵𝑊 → (𝐵 ∈ {𝐶, 𝐷} ↔ (𝐵 = 𝐶𝐵 = 𝐷)))
42, 3bi2anan9 637 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
51, 4bitr3d 281 1 ((𝐴𝑉𝐵𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wo 846   = wceq 1537  wcel 2108  wss 3976  {cpr 4650
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490  df-un 3981  df-ss 3993  df-sn 4649  df-pr 4651
This theorem is referenced by:  ssprsseq  4850
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