Theorem preleq 9623

Description: Equality of two unordered pairs when one member of each pair contains the other member. (Contributed by NM, 16-Oct-1996.) (Revised by AV, 15-Jun-2022.)

Hypothesis

Ref	Expression
preleq.b	⊢ 𝐵 ∈ V

Assertion

Ref	Expression
preleq	⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

Proof of Theorem preleq

Step	Hyp	Ref	Expression
1		preleq.b	. 2 ⊢ 𝐵 ∈ V
2		preleqg 9622	. 2 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ V ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))
3	1, 2	mp3anl2 1457	1 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1539   ∈ wcel 2107  Vcvv 3457  {cpr 4601

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-10 2140  ax-12 2176  ax-ext 2706  ax-sep 5264  ax-nul 5274  ax-pr 5400  ax-reg 9599

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-ne 2932  df-ral 3051  df-rex 3060  df-rab 3414  df-v 3459  df-dif 3927  df-un 3929  df-in 3931  df-ss 3941  df-nul 4307  df-if 4499  df-pw 4575  df-sn 4600  df-pr 4602  df-op 4606  df-br 5118  df-opab 5180  df-eprel 5551  df-fr 5604

This theorem is referenced by:  opthreg  9625  dfac2b  10138