Metamath Proof Explorer


Theorem brsuccf

Description: Binary relation form of the Succ function. (Contributed by Scott Fenton, 14-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Hypotheses brsuccf.1 AV
brsuccf.2 BV
Assertion brsuccf A𝖲𝗎𝖼𝖼BB=sucA

Proof

Step Hyp Ref Expression
1 brsuccf.1 AV
2 brsuccf.2 BV
3 df-succf 𝖲𝗎𝖼𝖼=𝖢𝗎𝗉I𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇
4 3 breqi A𝖲𝗎𝖼𝖼BA𝖢𝗎𝗉I𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇B
5 1 2 brco A𝖢𝗎𝗉I𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇BxAI𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇xx𝖢𝗎𝗉B
6 opex AAV
7 breq1 x=AAx𝖢𝗎𝗉BAA𝖢𝗎𝗉B
8 6 7 ceqsexv xx=AAx𝖢𝗎𝗉BAA𝖢𝗎𝗉B
9 snex AV
10 1 9 2 brcup AA𝖢𝗎𝗉BB=AA
11 8 10 bitri xx=AAx𝖢𝗎𝗉BB=AA
12 1 brtxp2 AI𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇xabx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇b
13 12 anbi1i AI𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇xx𝖢𝗎𝗉Babx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉B
14 3anass x=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇b
15 14 anbi1i x=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉Bx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉B
16 an32 x=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉Bx=abx𝖢𝗎𝗉BAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇b
17 vex aV
18 17 ideq AIaA=a
19 eqcom A=aa=A
20 18 19 bitri AIaa=A
21 vex bV
22 1 21 brsingle A𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bb=A
23 20 22 anbi12i AIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇ba=Ab=A
24 23 anbi1i AIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx=abx𝖢𝗎𝗉Ba=Ab=Ax=abx𝖢𝗎𝗉B
25 ancom x=abx𝖢𝗎𝗉BAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx=abx𝖢𝗎𝗉B
26 df-3an a=Ab=Ax=abx𝖢𝗎𝗉Ba=Ab=Ax=abx𝖢𝗎𝗉B
27 24 25 26 3bitr4i x=abx𝖢𝗎𝗉BAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇ba=Ab=Ax=abx𝖢𝗎𝗉B
28 15 16 27 3bitri x=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉Ba=Ab=Ax=abx𝖢𝗎𝗉B
29 28 2exbii abx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉Baba=Ab=Ax=abx𝖢𝗎𝗉B
30 19.41vv abx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉Babx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉B
31 opeq1 a=Aab=Ab
32 31 eqeq2d a=Ax=abx=Ab
33 32 anbi1d a=Ax=abx𝖢𝗎𝗉Bx=Abx𝖢𝗎𝗉B
34 opeq2 b=AAb=AA
35 34 eqeq2d b=Ax=Abx=AA
36 35 anbi1d b=Ax=Abx𝖢𝗎𝗉Bx=AAx𝖢𝗎𝗉B
37 1 9 33 36 ceqsex2v aba=Ab=Ax=abx𝖢𝗎𝗉Bx=AAx𝖢𝗎𝗉B
38 29 30 37 3bitr3i abx=abAIaA𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇bx𝖢𝗎𝗉Bx=AAx𝖢𝗎𝗉B
39 13 38 bitri AI𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇xx𝖢𝗎𝗉Bx=AAx𝖢𝗎𝗉B
40 39 exbii xAI𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇xx𝖢𝗎𝗉Bxx=AAx𝖢𝗎𝗉B
41 df-suc sucA=AA
42 41 eqeq2i B=sucAB=AA
43 11 40 42 3bitr4i xAI𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇xx𝖢𝗎𝗉BB=sucA
44 4 5 43 3bitri A𝖲𝗎𝖼𝖼BB=sucA