Metamath Proof Explorer


Theorem pcdiv

Description: Division property of the prime power function. (Contributed by Mario Carneiro, 1-Mar-2014)

Ref Expression
Assertion pcdiv P A A 0 B P pCnt A B = P pCnt A P pCnt B

Proof

Step Hyp Ref Expression
1 simp1 P A A 0 B P
2 simp2l P A A 0 B A
3 simp3 P A A 0 B B
4 znq A B A B
5 2 3 4 syl2anc P A A 0 B A B
6 2 zcnd P A A 0 B A
7 3 nncnd P A A 0 B B
8 simp2r P A A 0 B A 0
9 3 nnne0d P A A 0 B B 0
10 6 7 8 9 divne0d P A A 0 B A B 0
11 eqid sup n 0 | P n x < = sup n 0 | P n x <
12 eqid sup n 0 | P n y < = sup n 0 | P n y <
13 11 12 pcval P A B A B 0 P pCnt A B = ι z | x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y <
14 1 5 10 13 syl12anc P A A 0 B P pCnt A B = ι z | x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y <
15 eqid sup n 0 | P n A < = sup n 0 | P n A <
16 15 pczpre P A A 0 P pCnt A = sup n 0 | P n A <
17 16 3adant3 P A A 0 B P pCnt A = sup n 0 | P n A <
18 nnz B B
19 nnne0 B B 0
20 18 19 jca B B B 0
21 eqid sup n 0 | P n B < = sup n 0 | P n B <
22 21 pczpre P B B 0 P pCnt B = sup n 0 | P n B <
23 20 22 sylan2 P B P pCnt B = sup n 0 | P n B <
24 23 3adant2 P A A 0 B P pCnt B = sup n 0 | P n B <
25 17 24 oveq12d P A A 0 B P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n B <
26 eqid A B = A B
27 25 26 jctil P A A 0 B A B = A B P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n B <
28 oveq1 x = A x y = A y
29 28 eqeq2d x = A A B = x y A B = A y
30 breq2 x = A P n x P n A
31 30 rabbidv x = A n 0 | P n x = n 0 | P n A
32 31 supeq1d x = A sup n 0 | P n x < = sup n 0 | P n A <
33 32 oveq1d x = A sup n 0 | P n x < sup n 0 | P n y < = sup n 0 | P n A < sup n 0 | P n y <
34 33 eqeq2d x = A P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y < P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n y <
35 29 34 anbi12d x = A A B = x y P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y < A B = A y P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n y <
36 oveq2 y = B A y = A B
37 36 eqeq2d y = B A B = A y A B = A B
38 breq2 y = B P n y P n B
39 38 rabbidv y = B n 0 | P n y = n 0 | P n B
40 39 supeq1d y = B sup n 0 | P n y < = sup n 0 | P n B <
41 40 oveq2d y = B sup n 0 | P n A < sup n 0 | P n y < = sup n 0 | P n A < sup n 0 | P n B <
42 41 eqeq2d y = B P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n y < P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n B <
43 37 42 anbi12d y = B A B = A y P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n y < A B = A B P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n B <
44 35 43 rspc2ev A B A B = A B P pCnt A P pCnt B = sup n 0 | P n A < sup n 0 | P n B < x y A B = x y P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y <
45 2 3 27 44 syl3anc P A A 0 B x y A B = x y P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y <
46 ovex P pCnt A P pCnt B V
47 11 12 pceu P A B A B 0 ∃! z x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y <
48 1 5 10 47 syl12anc P A A 0 B ∃! z x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y <
49 eqeq1 z = P pCnt A P pCnt B z = sup n 0 | P n x < sup n 0 | P n y < P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y <
50 49 anbi2d z = P pCnt A P pCnt B A B = x y z = sup n 0 | P n x < sup n 0 | P n y < A B = x y P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y <
51 50 2rexbidv z = P pCnt A P pCnt B x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y < x y A B = x y P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y <
52 51 iota2 P pCnt A P pCnt B V ∃! z x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y < x y A B = x y P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y < ι z | x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y < = P pCnt A P pCnt B
53 46 48 52 sylancr P A A 0 B x y A B = x y P pCnt A P pCnt B = sup n 0 | P n x < sup n 0 | P n y < ι z | x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y < = P pCnt A P pCnt B
54 45 53 mpbid P A A 0 B ι z | x y A B = x y z = sup n 0 | P n x < sup n 0 | P n y < = P pCnt A P pCnt B
55 14 54 eqtrd P A A 0 B P pCnt A B = P pCnt A P pCnt B