Metamath Proof Explorer


Theorem finxp00

Description: Cartesian exponentiation of the empty set to any power is the empty set. (Contributed by ML, 24-Oct-2020)

Ref Expression
Assertion finxp00
|- ( (/) ^^ N ) = (/)

Proof

Step Hyp Ref Expression
1 finxpeq2
 |-  ( n = (/) -> ( (/) ^^ n ) = ( (/) ^^ (/) ) )
2 1 eqeq1d
 |-  ( n = (/) -> ( ( (/) ^^ n ) = (/) <-> ( (/) ^^ (/) ) = (/) ) )
3 finxpeq2
 |-  ( n = m -> ( (/) ^^ n ) = ( (/) ^^ m ) )
4 3 eqeq1d
 |-  ( n = m -> ( ( (/) ^^ n ) = (/) <-> ( (/) ^^ m ) = (/) ) )
5 finxpeq2
 |-  ( n = suc m -> ( (/) ^^ n ) = ( (/) ^^ suc m ) )
6 5 eqeq1d
 |-  ( n = suc m -> ( ( (/) ^^ n ) = (/) <-> ( (/) ^^ suc m ) = (/) ) )
7 finxpeq2
 |-  ( n = N -> ( (/) ^^ n ) = ( (/) ^^ N ) )
8 7 eqeq1d
 |-  ( n = N -> ( ( (/) ^^ n ) = (/) <-> ( (/) ^^ N ) = (/) ) )
9 finxp0
 |-  ( (/) ^^ (/) ) = (/)
10 suceq
 |-  ( m = (/) -> suc m = suc (/) )
11 df-1o
 |-  1o = suc (/)
12 10 11 eqtr4di
 |-  ( m = (/) -> suc m = 1o )
13 finxpeq2
 |-  ( suc m = 1o -> ( (/) ^^ suc m ) = ( (/) ^^ 1o ) )
14 12 13 syl
 |-  ( m = (/) -> ( (/) ^^ suc m ) = ( (/) ^^ 1o ) )
15 finxp1o
 |-  ( (/) ^^ 1o ) = (/)
16 14 15 eqtrdi
 |-  ( m = (/) -> ( (/) ^^ suc m ) = (/) )
17 16 adantl
 |-  ( ( m e. _om /\ m = (/) ) -> ( (/) ^^ suc m ) = (/) )
18 finxpsuc
 |-  ( ( m e. _om /\ m =/= (/) ) -> ( (/) ^^ suc m ) = ( ( (/) ^^ m ) X. (/) ) )
19 xp0
 |-  ( ( (/) ^^ m ) X. (/) ) = (/)
20 18 19 eqtrdi
 |-  ( ( m e. _om /\ m =/= (/) ) -> ( (/) ^^ suc m ) = (/) )
21 17 20 pm2.61dane
 |-  ( m e. _om -> ( (/) ^^ suc m ) = (/) )
22 21 a1d
 |-  ( m e. _om -> ( ( (/) ^^ m ) = (/) -> ( (/) ^^ suc m ) = (/) ) )
23 2 4 6 8 9 22 finds
 |-  ( N e. _om -> ( (/) ^^ N ) = (/) )
24 finxpnom
 |-  ( -. N e. _om -> ( (/) ^^ N ) = (/) )
25 23 24 pm2.61i
 |-  ( (/) ^^ N ) = (/)