Metamath Proof Explorer


Theorem 3factsumint

Description: Helpful equation for lcm inequality proof. (Contributed by metakunt, 26-Apr-2024)

Ref Expression
Hypotheses 3factsumint.1 A=LU
3factsumint.2 φBFin
3factsumint.3 φL
3factsumint.4 φU
3factsumint.5 φxAF:Acn
3factsumint.6 φkBG
3factsumint.7 φkBxAH:Acn
Assertion 3factsumint φAFkBGHdx=kBGAFHdx

Proof

Step Hyp Ref Expression
1 3factsumint.1 A=LU
2 3factsumint.2 φBFin
3 3factsumint.3 φL
4 3factsumint.4 φU
5 3factsumint.5 φxAF:Acn
6 3factsumint.6 φkBG
7 3factsumint.7 φkBxAH:Acn
8 cncff xAF:AcnxAF:A
9 5 8 syl φxAF:A
10 eqid xAF=xAF
11 10 fmpt xAFxAF:A
12 9 11 sylibr φxAF
13 12 r19.21bi φxAF
14 cncff xAH:AcnxAH:A
15 7 14 syl φkBxAH:A
16 eqid xAH=xAH
17 16 fmpt xAHxAH:A
18 15 17 sylibr φkBxAH
19 18 r19.21bi φkBxAH
20 anass φkBxAφkBxA
21 ancom kBxAxAkB
22 21 anbi2i φkBxAφxAkB
23 20 22 bitri φkBxAφxAkB
24 23 imbi1i φkBxAHφxAkBH
25 19 24 mpbi φxAkBH
26 2 13 6 25 3factsumint4 φAkBFGHdx=AFkBGHdx
27 1 2 3 4 13 5 6 25 7 3factsumint1 φAkBFGHdx=kBAFGHdx
28 26 27 eqtr3d φAFkBGHdx=kBAFGHdx
29 13 6 25 3factsumint2 φkBAFGHdx=kBAGFHdx
30 1 3 4 13 5 6 25 7 3factsumint3 φkBAGFHdx=kBGAFHdx
31 28 29 30 3eqtrd φAFkBGHdx=kBGAFHdx