Theorem ssprss 4784

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 4779	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ {𝐴, 𝐵} ⊆ {𝐶, 𝐷}))
2		elprg 4606	. . 3 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∨ 𝐴 = 𝐷)))
3		elprg 4606	. . 3 ⊢ (𝐵 ∈ 𝑊 → (𝐵 ∈ {𝐶, 𝐷} ↔ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷)))
4	2, 3	bi2anan9 650	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ((𝐴 ∈ {𝐶, 𝐷} ∧ 𝐵 ∈ {𝐶, 𝐷}) ↔ ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷))))
5	1, 4	bitr3d 284	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷))))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1569   ∈ wcel 2144   ⊆ wss 3895  {cpr 4585

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3896  df-un 4151  df-sn 4584  df-pr 4586

This theorem is used by:  ssprsseq  4785