Metamath Proof Explorer


Theorem ixxdisj

Description: Split an interval into disjoint pieces. (Contributed by Mario Carneiro, 16-Jun-2014)

Ref Expression
Hypotheses ixx.1 O=x*,y*z*|xRzzSy
ixxun.2 P=x*,y*z*|xTzzUy
ixxun.3 B*w*BTw¬wSB
Assertion ixxdisj A*B*C*AOBBPC=

Proof

Step Hyp Ref Expression
1 ixx.1 O=x*,y*z*|xRzzSy
2 ixxun.2 P=x*,y*z*|xTzzUy
3 ixxun.3 B*w*BTw¬wSB
4 elin wAOBBPCwAOBwBPC
5 1 elixx1 A*B*wAOBw*ARwwSB
6 5 3adant3 A*B*C*wAOBw*ARwwSB
7 6 biimpa A*B*C*wAOBw*ARwwSB
8 7 simp3d A*B*C*wAOBwSB
9 8 adantrr A*B*C*wAOBwBPCwSB
10 2 elixx1 B*C*wBPCw*BTwwUC
11 10 3adant1 A*B*C*wBPCw*BTwwUC
12 11 biimpa A*B*C*wBPCw*BTwwUC
13 12 simp2d A*B*C*wBPCBTw
14 simpl2 A*B*C*wBPCB*
15 12 simp1d A*B*C*wBPCw*
16 14 15 3 syl2anc A*B*C*wBPCBTw¬wSB
17 13 16 mpbid A*B*C*wBPC¬wSB
18 17 adantrl A*B*C*wAOBwBPC¬wSB
19 9 18 pm2.65da A*B*C*¬wAOBwBPC
20 19 pm2.21d A*B*C*wAOBwBPCw
21 4 20 biimtrid A*B*C*wAOBBPCw
22 21 ssrdv A*B*C*AOBBPC
23 ss0 AOBBPCAOBBPC=
24 22 23 syl A*B*C*AOBBPC=