Metamath Proof Explorer


Theorem disjdmqscossss

Description: Lemma for disjdmqseq via disjdmqs . (Contributed by Peter Mazsa, 16-Sep-2021)

Ref Expression
Assertion disjdmqscossss Disj R dom R / R dom R / R

Proof

Step Hyp Ref Expression
1 disjrel Disj R Rel R
2 releldmqscoss v V Rel R v dom R / R u dom R x u R v = x R
3 2 elv Rel R v dom R / R u dom R x u R v = x R
4 1 3 syl Disj R v dom R / R u dom R x u R v = x R
5 disjlem19 x V Disj R u dom R x u R u R = x R
6 5 elv Disj R u dom R x u R u R = x R
7 6 ralrimivv Disj R u dom R x u R u R = x R
8 2r19.29 u dom R x u R u R = x R u dom R x u R v = x R u dom R x u R u R = x R v = x R
9 8 ex u dom R x u R u R = x R u dom R x u R v = x R u dom R x u R u R = x R v = x R
10 7 9 syl Disj R u dom R x u R v = x R u dom R x u R u R = x R v = x R
11 4 10 sylbid Disj R v dom R / R u dom R x u R u R = x R v = x R
12 eqtr3 u R = x R v = x R u R = v
13 12 reximi x u R u R = x R v = x R x u R u R = v
14 13 reximi u dom R x u R u R = x R v = x R u dom R x u R u R = v
15 11 14 syl6 Disj R v dom R / R u dom R x u R u R = v
16 df-rex x u R u R = v x x u R u R = v
17 19.41v x x u R u R = v x x u R u R = v
18 16 17 bitri x u R u R = v x x u R u R = v
19 18 simprbi x u R u R = v u R = v
20 19 reximi u dom R x u R u R = v u dom R u R = v
21 15 20 syl6 Disj R v dom R / R u dom R u R = v
22 eqcom u R = v v = u R
23 22 rexbii u dom R u R = v u dom R v = u R
24 21 23 imbitrdi Disj R v dom R / R u dom R v = u R
25 24 ss2abdv Disj R v | v dom R / R v | u dom R v = u R
26 abid1 dom R / R = v | v dom R / R
27 df-qs dom R / R = v | u dom R v = u R
28 25 26 27 3sstr4g Disj R dom R / R dom R / R