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 A V
brsuccf.2 B V
Assertion brsuccf A 𝖲𝗎𝖼𝖼 B B = suc A

Proof

Step Hyp Ref Expression
1 brsuccf.1 A V
2 brsuccf.2 B V
3 df-succf 𝖲𝗎𝖼𝖼 = 𝖢𝗎𝗉 I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇
4 3 breqi A 𝖲𝗎𝖼𝖼 B A 𝖢𝗎𝗉 I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 B
5 1 2 brco A 𝖢𝗎𝗉 I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 B x A I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x x 𝖢𝗎𝗉 B
6 opex A A V
7 breq1 x = A A x 𝖢𝗎𝗉 B A A 𝖢𝗎𝗉 B
8 6 7 ceqsexv x x = A A x 𝖢𝗎𝗉 B A A 𝖢𝗎𝗉 B
9 snex A V
10 1 9 2 brcup A A 𝖢𝗎𝗉 B B = A A
11 8 10 bitri x x = A A x 𝖢𝗎𝗉 B B = A A
12 1 brtxp2 A I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x a b x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b
13 12 anbi1i A I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x x 𝖢𝗎𝗉 B a b x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B
14 3anass x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b
15 14 anbi1i x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B
16 an32 x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B x = a b x 𝖢𝗎𝗉 B A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b
17 vex a V
18 17 ideq A I a A = a
19 eqcom A = a a = A
20 18 19 bitri A I a a = A
21 vex b V
22 1 21 brsingle A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b b = A
23 20 22 anbi12i A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b a = A b = A
24 23 anbi1i A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x = a b x 𝖢𝗎𝗉 B a = A b = A x = a b x 𝖢𝗎𝗉 B
25 ancom x = a b x 𝖢𝗎𝗉 B A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x = a b x 𝖢𝗎𝗉 B
26 df-3an a = A b = A x = a b x 𝖢𝗎𝗉 B a = A b = A x = a b x 𝖢𝗎𝗉 B
27 24 25 26 3bitr4i x = a b x 𝖢𝗎𝗉 B A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b a = A b = A x = a b x 𝖢𝗎𝗉 B
28 15 16 27 3bitri x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B a = A b = A x = a b x 𝖢𝗎𝗉 B
29 28 2exbii a b x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B a b a = A b = A x = a b x 𝖢𝗎𝗉 B
30 19.41vv a b x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B a b x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B
31 opeq1 a = A a b = A b
32 31 eqeq2d a = A x = a b x = A b
33 32 anbi1d a = A x = a b x 𝖢𝗎𝗉 B x = A b x 𝖢𝗎𝗉 B
34 opeq2 b = A A b = A A
35 34 eqeq2d b = A x = A b x = A A
36 35 anbi1d b = A x = A b x 𝖢𝗎𝗉 B x = A A x 𝖢𝗎𝗉 B
37 1 9 33 36 ceqsex2v a b a = A b = A x = a b x 𝖢𝗎𝗉 B x = A A x 𝖢𝗎𝗉 B
38 29 30 37 3bitr3i a b x = a b A I a A 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 b x 𝖢𝗎𝗉 B x = A A x 𝖢𝗎𝗉 B
39 13 38 bitri A I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x x 𝖢𝗎𝗉 B x = A A x 𝖢𝗎𝗉 B
40 39 exbii x A I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x x 𝖢𝗎𝗉 B x x = A A x 𝖢𝗎𝗉 B
41 df-suc suc A = A A
42 41 eqeq2i B = suc A B = A A
43 11 40 42 3bitr4i x A I 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x x 𝖢𝗎𝗉 B B = suc A
44 4 5 43 3bitri A 𝖲𝗎𝖼𝖼 B B = suc A