Metamath Proof Explorer


Theorem efsep

Description: Separate out the next term of the power series expansion of the exponential function. The last hypothesis allows the separated terms to be rearranged as desired. (Contributed by Paul Chapman, 23-Nov-2007) (Revised by Mario Carneiro, 29-Apr-2014)

Ref Expression
Hypotheses efsep.1 F=n0Ann!
efsep.2 N=M+1
efsep.3 M0
efsep.4 φA
efsep.5 φB
efsep.6 φeA=B+kMFk
efsep.7 φB+AMM!=D
Assertion efsep φeA=D+kNFk

Proof

Step Hyp Ref Expression
1 efsep.1 F=n0Ann!
2 efsep.2 N=M+1
3 efsep.3 M0
4 efsep.4 φA
5 efsep.5 φB
6 efsep.6 φeA=B+kMFk
7 efsep.7 φB+AMM!=D
8 eqid M=M
9 3 nn0zi M
10 9 a1i φM
11 eqidd φkMFk=Fk
12 eluznn0 M0kMk0
13 3 12 mpan kMk0
14 1 eftval k0Fk=Akk!
15 14 adantl φk0Fk=Akk!
16 eftcl Ak0Akk!
17 4 16 sylan φk0Akk!
18 15 17 eqeltrd φk0Fk
19 13 18 sylan2 φkMFk
20 1 eftlcvg AM0seqM+Fdom
21 4 3 20 sylancl φseqM+Fdom
22 8 10 11 19 21 isum1p φkMFk=FM+kM+1Fk
23 1 eftval M0FM=AMM!
24 3 23 ax-mp FM=AMM!
25 2 eqcomi M+1=N
26 25 fveq2i M+1=N
27 26 sumeq1i kM+1Fk=kNFk
28 24 27 oveq12i FM+kM+1Fk=AMM!+kNFk
29 22 28 eqtrdi φkMFk=AMM!+kNFk
30 29 oveq2d φB+kMFk=B+AMM!+kNFk
31 eftcl AM0AMM!
32 4 3 31 sylancl φAMM!
33 peano2nn0 M0M+10
34 3 33 ax-mp M+10
35 2 34 eqeltri N0
36 1 eftlcl AN0kNFk
37 4 35 36 sylancl φkNFk
38 5 32 37 addassd φB+AMM!+kNFk=B+AMM!+kNFk
39 30 38 eqtr4d φB+kMFk=B+AMM!+kNFk
40 7 oveq1d φB+AMM!+kNFk=D+kNFk
41 6 39 40 3eqtrd φeA=D+kNFk