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Theorem ssprss 4723
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 4718 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ {𝐴, 𝐵} ⊆ {𝐶, 𝐷}))
2 elprg 4548 . . 3 (𝐴𝑉 → (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐴 = 𝐷)))
3 elprg 4548 . . 3 (𝐵𝑊 → (𝐵 ∈ {𝐶, 𝐷} ↔ (𝐵 = 𝐶𝐵 = 𝐷)))
42, 3bi2anan9 639 . 2 ((𝐴𝑉𝐵𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
51, 4bitr3d 284 1 ((𝐴𝑉𝐵𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐴 = 𝐷) ∧ (𝐵 = 𝐶𝐵 = 𝐷))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 399  wo 847   = wceq 1543  wcel 2112  wss 3853  {cpr 4529
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-ext 2708
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-tru 1546  df-ex 1788  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809  df-v 3400  df-un 3858  df-in 3860  df-ss 3870  df-sn 4528  df-pr 4530
This theorem is referenced by:  ssprsseq  4724
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