Metamath Proof Explorer

Theorem ackval1012

Description: The Ackermann function at (1,0), (1,1), (1,2). (Contributed by AV, 4-May-2024)

Ref Expression
Assertion ackval1012 
|- <. ( ( Ack ` 1 ) ` 0 ) , ( ( Ack ` 1 ) ` 1 ) , ( ( Ack ` 1 ) ` 2 ) >. = <. 2 , 3 , 4 >.

Proof

Step Hyp Ref Expression
1 ackval1 
 |-  ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) )
2 oveq1 
 |-  ( n = 0 -> ( n + 2 ) = ( 0 + 2 ) )
3 2cn 
 |-  2 e. CC
4 3 addid2i 
 |-  ( 0 + 2 ) = 2
5 2 4 eqtrdi 
 |-  ( n = 0 -> ( n + 2 ) = 2 )
6 0nn0 
 |-  0 e. NN0
7 6 a1i 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> 0 e. NN0 )
8 2nn0 
 |-  2 e. NN0
9 8 a1i 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> 2 e. NN0 )
10 1 5 7 9 fvmptd3 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> ( ( Ack ` 1 ) ` 0 ) = 2 )
11 oveq1 
 |-  ( n = 1 -> ( n + 2 ) = ( 1 + 2 ) )
12 1p2e3 
 |-  ( 1 + 2 ) = 3
13 11 12 eqtrdi 
 |-  ( n = 1 -> ( n + 2 ) = 3 )
14 1nn0 
 |-  1 e. NN0
15 14 a1i 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> 1 e. NN0 )
16 3nn0 
 |-  3 e. NN0
17 16 a1i 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> 3 e. NN0 )
18 1 13 15 17 fvmptd3 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> ( ( Ack ` 1 ) ` 1 ) = 3 )
19 oveq1 
 |-  ( n = 2 -> ( n + 2 ) = ( 2 + 2 ) )
20 2p2e4 
 |-  ( 2 + 2 ) = 4
21 19 20 eqtrdi 
 |-  ( n = 2 -> ( n + 2 ) = 4 )
22 4nn0 
 |-  4 e. NN0
23 22 a1i 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> 4 e. NN0 )
24 1 21 9 23 fvmptd3 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> ( ( Ack ` 1 ) ` 2 ) = 4 )
25 10 18 24 oteq123d 
 |-  ( ( Ack ` 1 ) = ( n e. NN0 |-> ( n + 2 ) ) -> <. ( ( Ack ` 1 ) ` 0 ) , ( ( Ack ` 1 ) ` 1 ) , ( ( Ack ` 1 ) ` 2 ) >. = <. 2 , 3 , 4 >. )
26 1 25 ax-mp 
 |-  <. ( ( Ack ` 1 ) ` 0 ) , ( ( Ack ` 1 ) ` 1 ) , ( ( Ack ` 1 ) ` 2 ) >. = <. 2 , 3 , 4 >.