Metamath Proof Explorer


Theorem funpartfv

Description: The function value of the functional part is identical to the original functional value. (Contributed by Scott Fenton, 17-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion funpartfv 𝖥𝗎𝗇𝖯𝖺𝗋𝗍 F A = F A

Proof

Step Hyp Ref Expression
1 df-funpart 𝖥𝗎𝗇𝖯𝖺𝗋𝗍 F = F dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌
2 1 fveq1i 𝖥𝗎𝗇𝖯𝖺𝗋𝗍 F A = F dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 A
3 fvres A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 F dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 A = F A
4 nfvres ¬ A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 F dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 A =
5 funpartlem A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 x F A = x
6 eusn ∃! x x F A x F A = x
7 5 6 bitr4i A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 ∃! x x F A
8 elimasng A V x V x F A A x F
9 8 elvd A V x F A A x F
10 df-br A F x A x F
11 9 10 bitr4di A V x F A A F x
12 11 eubidv A V ∃! x x F A ∃! x A F x
13 7 12 syl5bb A V A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 ∃! x A F x
14 13 notbid A V ¬ A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 ¬ ∃! x A F x
15 tz6.12-2 ¬ ∃! x A F x F A =
16 14 15 syl6bi A V ¬ A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 F A =
17 fvprc ¬ A V F A =
18 17 a1d ¬ A V ¬ A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 F A =
19 16 18 pm2.61i ¬ A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 F A =
20 4 19 eqtr4d ¬ A dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 F dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 A = F A
21 3 20 pm2.61i F dom 𝖨𝗆𝖺𝗀𝖾 F 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 V × 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇𝗌 A = F A
22 2 21 eqtri 𝖥𝗎𝗇𝖯𝖺𝗋𝗍 F A = F A