Metamath Proof Explorer


Theorem dmqsblocks

Description: If the pet span ( R |X. ( ' E | A ) ) partitions A , then every block u e. A is of the form [ v ] for some v that not only lies in the domain but also has at least one internal element c and at least one R -target b (cf. also the comments of qseq ). It makes explicit that pet gives active representatives for each block, without ever forcing v = u . (Contributed by Peter Mazsa, 23-Nov-2025)

Ref Expression
Assertion dmqsblocks dom R E -1 A / R E -1 A = A u A v dom R E -1 A b c u = v R E -1 A c v v R b

Proof

Step Hyp Ref Expression
1 qseq dom R E -1 A / R E -1 A = A u u A v dom R E -1 A u = v R E -1 A
2 eqab2 u u A v dom R E -1 A u = v R E -1 A u A v dom R E -1 A u = v R E -1 A
3 1 2 sylbi dom R E -1 A / R E -1 A = A u A v dom R E -1 A u = v R E -1 A
4 rexanid v dom R E -1 A v dom R E -1 A u = v R E -1 A v dom R E -1 A u = v R E -1 A
5 eldmxrncnvepres2 v V v dom R E -1 A v A c c v b v R b
6 5 elv v dom R E -1 A v A c c v b v R b
7 3simpc v A c c v b v R b c c v b v R b
8 6 7 sylbi v dom R E -1 A c c v b v R b
9 exdistrv c b c v v R b c c v b v R b
10 excom c b c v v R b b c c v v R b
11 9 10 bitr3i c c v b v R b b c c v v R b
12 8 11 sylib v dom R E -1 A b c c v v R b
13 12 anim1ci v dom R E -1 A u = v R E -1 A u = v R E -1 A b c c v v R b
14 3anass u = v R E -1 A c v v R b u = v R E -1 A c v v R b
15 14 2exbii b c u = v R E -1 A c v v R b b c u = v R E -1 A c v v R b
16 19.42vv b c u = v R E -1 A c v v R b u = v R E -1 A b c c v v R b
17 15 16 sylbbr u = v R E -1 A b c c v v R b b c u = v R E -1 A c v v R b
18 13 17 syl v dom R E -1 A u = v R E -1 A b c u = v R E -1 A c v v R b
19 18 reximi v dom R E -1 A v dom R E -1 A u = v R E -1 A v dom R E -1 A b c u = v R E -1 A c v v R b
20 4 19 sylbir v dom R E -1 A u = v R E -1 A v dom R E -1 A b c u = v R E -1 A c v v R b
21 20 ralimi u A v dom R E -1 A u = v R E -1 A u A v dom R E -1 A b c u = v R E -1 A c v v R b
22 3 21 syl dom R E -1 A / R E -1 A = A u A v dom R E -1 A b c u = v R E -1 A c v v R b