Theorem brvbrvvdif 38950

Description: Binary relation with the complement under the universal class of ordered pairs is the same as with universal complement. (Contributed by Peter Mazsa, 28-Nov-2018.)

Assertion

Ref	Expression
brvbrvvdif	⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴((V × V) ∖ 𝑅)𝐵 ↔ 𝐴(V ∖ 𝑅)𝐵))

Proof of Theorem brvbrvvdif

Step	Hyp	Ref	Expression
1		brvvdif 38949	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴((V × V) ∖ 𝑅)𝐵 ↔ ¬ 𝐴𝑅𝐵))
2		brvdif 38947	. 2 ⊢ (𝐴(V ∖ 𝑅)𝐵 ↔ ¬ 𝐴𝑅𝐵)
3	1, 2	bitr4di 292	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴((V × V) ∖ 𝑅)𝐵 ↔ 𝐴(V ∖ 𝑅)𝐵))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∈ wcel 2146  Vcvv 3458   ∖ cdif 3905   class class class wbr 5112   × cxp 5662

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738  ax-sep 5260  ax-pr 5407

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4491  df-sn 4593  df-pr 4595  df-op 4599  df-br 5113  df-opab 5177  df-xp 5670

This theorem is used by: (None)