Theorem brvbrvvdif 36330

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 36329	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴((V × V) ∖ 𝑅)𝐵 ↔ ¬ 𝐴𝑅𝐵))
2		brvdif 36327	. 2 ⊢ (𝐴(V ∖ 𝑅)𝐵 ↔ ¬ 𝐴𝑅𝐵)
3	1, 2	bitr4di 288	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴((V × V) ∖ 𝑅)𝐵 ↔ 𝐴(V ∖ 𝑅)𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395   ∈ wcel 2108  Vcvv 3422   ∖ cdif 3880   class class class wbr 5070   × cxp 5578

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-opab 5133  df-xp 5586

This theorem is referenced by: (None)