Metamath Proof Explorer


Theorem sumeq1

Description: Equality theorem for a sum. (Contributed by NM, 11-Dec-2005) (Revised by Mario Carneiro, 13-Jun-2019)

Ref Expression
Assertion sumeq1 A = B k A C = k B C

Proof

Step Hyp Ref Expression
1 sseq1 A = B A m B m
2 simpl A = B n A = B
3 2 eleq2d A = B n n A n B
4 3 ifbid A = B n if n A n / k C 0 = if n B n / k C 0
5 4 mpteq2dva A = B n if n A n / k C 0 = n if n B n / k C 0
6 5 seqeq3d A = B seq m + n if n A n / k C 0 = seq m + n if n B n / k C 0
7 6 breq1d A = B seq m + n if n A n / k C 0 x seq m + n if n B n / k C 0 x
8 1 7 anbi12d A = B A m seq m + n if n A n / k C 0 x B m seq m + n if n B n / k C 0 x
9 8 rexbidv A = B m A m seq m + n if n A n / k C 0 x m B m seq m + n if n B n / k C 0 x
10 f1oeq3 A = B f : 1 m 1-1 onto A f : 1 m 1-1 onto B
11 10 anbi1d A = B f : 1 m 1-1 onto A x = seq 1 + n f n / k C m f : 1 m 1-1 onto B x = seq 1 + n f n / k C m
12 11 exbidv A = B f f : 1 m 1-1 onto A x = seq 1 + n f n / k C m f f : 1 m 1-1 onto B x = seq 1 + n f n / k C m
13 12 rexbidv A = B m f f : 1 m 1-1 onto A x = seq 1 + n f n / k C m m f f : 1 m 1-1 onto B x = seq 1 + n f n / k C m
14 9 13 orbi12d A = B 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 m B m seq m + n if n B n / k C 0 x m f f : 1 m 1-1 onto B x = seq 1 + n f n / k C m
15 14 iotabidv A = B 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 = ι x | m B m seq m + n if n B n / k C 0 x m f f : 1 m 1-1 onto B x = seq 1 + n f n / k C m
16 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
17 df-sum k B C = ι x | m B m seq m + n if n B n / k C 0 x m f f : 1 m 1-1 onto B x = seq 1 + n f n / k C m
18 15 16 17 3eqtr4g A = B k A C = k B C