Metamath Proof Explorer


Theorem cbvsum

Description: Change bound variable in a sum. (Contributed by NM, 11-Dec-2005) (Revised by Mario Carneiro, 13-Jun-2019)

Ref Expression
Hypotheses cbvsum.1 j = k B = C
cbvsum.2 _ k B
cbvsum.3 _ j C
Assertion cbvsum j A B = k A C

Proof

Step Hyp Ref Expression
1 cbvsum.1 j = k B = C
2 cbvsum.2 _ k B
3 cbvsum.3 _ j C
4 2 3 1 cbvcsbw n / j B = n / k C
5 4 a1i n / j B = n / k C
6 5 ifeq1d if n A n / j B 0 = if n A n / k C 0
7 6 mpteq2dv n if n A n / j B 0 = n if n A n / k C 0
8 7 seqeq3d seq m + n if n A n / j B 0 = seq m + n if n A n / k C 0
9 8 mptru seq m + n if n A n / j B 0 = seq m + n if n A n / k C 0
10 9 breq1i seq m + n if n A n / j B 0 x seq m + n if n A n / k C 0 x
11 10 anbi2i A m seq m + n if n A n / j B 0 x A m seq m + n if n A n / k C 0 x
12 11 rexbii m A m seq m + n if n A n / j B 0 x m A m seq m + n if n A n / k C 0 x
13 2 3 1 cbvcsbw f n / j B = f n / k C
14 13 a1i f n / j B = f n / k C
15 14 mpteq2dv n f n / j B = n f n / k C
16 15 seqeq3d seq 1 + n f n / j B = seq 1 + n f n / k C
17 16 mptru seq 1 + n f n / j B = seq 1 + n f n / k C
18 17 fveq1i seq 1 + n f n / j B m = seq 1 + n f n / k C m
19 18 eqeq2i x = seq 1 + n f n / j B m x = seq 1 + n f n / k C m
20 19 anbi2i f : 1 m 1-1 onto A x = seq 1 + n f n / j B m f : 1 m 1-1 onto A x = seq 1 + n f n / k C m
21 20 exbii f f : 1 m 1-1 onto A x = seq 1 + n f n / j B m f f : 1 m 1-1 onto A x = seq 1 + n f n / k C m
22 21 rexbii m f f : 1 m 1-1 onto A x = seq 1 + n f n / j B m m f f : 1 m 1-1 onto A x = seq 1 + n f n / k C m
23 12 22 orbi12i m A m seq m + n if n A n / j B 0 x m f f : 1 m 1-1 onto A x = seq 1 + n f n / j B m m A m seq m + n if n A n / k C 0 x m f f : 1 m 1-1 onto A x = seq 1 + n f n / k C m
24 23 iotabii ι x | m A m seq m + n if n A n / j B 0 x m f f : 1 m 1-1 onto A x = seq 1 + n f n / j B m = ι x | m A m seq m + n if n A n / k C 0 x m f f : 1 m 1-1 onto A x = seq 1 + n f n / k C m
25 df-sum j A B = ι x | m A m seq m + n if n A n / j B 0 x m f f : 1 m 1-1 onto A x = seq 1 + n f n / j B m
26 df-sum k A C = ι x | m A m seq m + n if n A n / k C 0 x m f f : 1 m 1-1 onto A x = seq 1 + n f n / k C m
27 24 25 26 3eqtr4i j A B = k A C