Metamath Proof Explorer


Theorem ditgswap

Description: Reverse a directed integral. (Contributed by Mario Carneiro, 13-Aug-2014)

Ref Expression
Hypotheses ditgcl.x φX
ditgcl.y φY
ditgcl.a φAXY
ditgcl.b φBXY
ditgcl.c φxXYCV
ditgcl.i φxXYC𝐿1
Assertion ditgswap φBACdx=ABCdx

Proof

Step Hyp Ref Expression
1 ditgcl.x φX
2 ditgcl.y φY
3 ditgcl.a φAXY
4 ditgcl.b φBXY
5 ditgcl.c φxXYCV
6 ditgcl.i φxXYC𝐿1
7 elicc2 XYAXYAXAAY
8 1 2 7 syl2anc φAXYAXAAY
9 3 8 mpbid φAXAAY
10 9 simp1d φA
11 elicc2 XYBXYBXBBY
12 1 2 11 syl2anc φBXYBXBBY
13 4 12 mpbid φBXBBY
14 13 simp1d φB
15 simpr φABAB
16 10 adantr φABA
17 14 adantr φABB
18 15 16 17 ditgneg φABBACdx=ABCdx
19 15 ditgpos φABABCdx=ABCdx
20 19 negeqd φABABCdx=ABCdx
21 18 20 eqtr4d φABBACdx=ABCdx
22 1 rexrd φX*
23 13 simp2d φXB
24 iooss1 X*XBBAXA
25 22 23 24 syl2anc φBAXA
26 2 rexrd φY*
27 9 simp3d φAY
28 iooss2 Y*AYXAXY
29 26 27 28 syl2anc φXAXY
30 25 29 sstrd φBAXY
31 30 sselda φxBAxXY
32 iblmbf xXYC𝐿1xXYCMblFn
33 6 32 syl φxXYCMblFn
34 33 5 mbfmptcl φxXYC
35 31 34 syldan φxBAC
36 ioombl BAdomvol
37 36 a1i φBAdomvol
38 30 37 5 6 iblss φxBAC𝐿1
39 35 38 itgcl φBACdx
40 39 adantr φBABACdx
41 40 negnegd φBABACdx=BACdx
42 simpr φBABA
43 14 adantr φBAB
44 10 adantr φBAA
45 42 43 44 ditgneg φBAABCdx=BACdx
46 45 negeqd φBAABCdx=BACdx
47 42 ditgpos φBABACdx=BACdx
48 41 46 47 3eqtr4rd φBABACdx=ABCdx
49 10 14 21 48 lecasei φBACdx=ABCdx