Metamath Proof Explorer


Theorem iocinico

Description: The intersection of two sets that meet at a point is that point. (Contributed by Jon Pennant, 12-Jun-2019)

Ref Expression
Assertion iocinico A * B * C * A < B B < C A B B C = B

Proof

Step Hyp Ref Expression
1 df-in A B B C = x | x A B x B C
2 elioc1 A * B * x A B x * A < x x B
3 2 3adant3 A * B * C * x A B x * A < x x B
4 3simpb x * A < x x B x * x B
5 3 4 syl6bi A * B * C * x A B x * x B
6 elico1 B * C * x B C x * B x x < C
7 6 3adant1 A * B * C * x B C x * B x x < C
8 3simpa x * B x x < C x * B x
9 7 8 syl6bi A * B * C * x B C x * B x
10 5 9 anim12d A * B * C * x A B x B C x * x B x * B x
11 simpll x * x B x * B x x *
12 simprr x * x B x * B x B x
13 simplr x * x B x * B x x B
14 11 12 13 3jca x * x B x * B x x * B x x B
15 10 14 syl6 A * B * C * x A B x B C x * B x x B
16 elicc1 B * B * x B B x * B x x B
17 16 anidms B * x B B x * B x x B
18 17 3ad2ant2 A * B * C * x B B x * B x x B
19 15 18 sylibrd A * B * C * x A B x B C x B B
20 19 ss2abdv A * B * C * x | x A B x B C x | x B B
21 1 20 eqsstrid A * B * C * A B B C x | x B B
22 abid2 x | x B B = B B
23 21 22 sseqtrdi A * B * C * A B B C B B
24 23 adantr A * B * C * A < B B < C A B B C B B
25 iccid B * B B = B
26 25 3ad2ant2 A * B * C * B B = B
27 26 adantr A * B * C * A < B B < C B B = B
28 24 27 sseqtrd A * B * C * A < B B < C A B B C B
29 simpl2 A * B * C * A < B B < C B *
30 simprl A * B * C * A < B B < C A < B
31 29 xrleidd A * B * C * A < B B < C B B
32 elioc1 A * B * B A B B * A < B B B
33 32 3adant3 A * B * C * B A B B * A < B B B
34 33 adantr A * B * C * A < B B < C B A B B * A < B B B
35 29 30 31 34 mpbir3and A * B * C * A < B B < C B A B
36 simprr A * B * C * A < B B < C B < C
37 elico1 B * C * B B C B * B B B < C
38 37 3adant1 A * B * C * B B C B * B B B < C
39 38 adantr A * B * C * A < B B < C B B C B * B B B < C
40 29 31 36 39 mpbir3and A * B * C * A < B B < C B B C
41 35 40 elind A * B * C * A < B B < C B A B B C
42 41 snssd A * B * C * A < B B < C B A B B C
43 28 42 eqssd A * B * C * A < B B < C A B B C = B