Theorem ssprss 4805

Description: A pair as subset of a pair. (Contributed by AV, 26-Oct-2020.)

Assertion

Ref	Expression
ssprss	⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷))))

Proof of Theorem ssprss

Step	Hyp	Ref	Expression
1		prssg 4800	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ {𝐴, 𝐵} ⊆ {𝐶, 𝐷}))
2		elprg 4629	. . 3 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∨ 𝐴 = 𝐷)))
3		elprg 4629	. . 3 ⊢ (𝐵 ∈ 𝑊 → (𝐵 ∈ {𝐶, 𝐷} ↔ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷)))
4	2, 3	bi2anan9 638	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷))))
5	1, 4	bitr3d 281	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷))))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∨ wo 847   = wceq 1540   ∈ wcel 2109   ⊆ wss 3931  {cpr 4608

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 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-v 3466  df-un 3936  df-ss 3948  df-sn 4607  df-pr 4609

This theorem is referenced by:  ssprsseq  4806