Metamath Proof Explorer


Theorem hashnnsuc

Description: The # function on _om turns successor into adding 1. (Contributed by Eric Schmidt, 7-Jul-2026)

Ref Expression
Assertion hashnnsuc
|- ( A e. _om -> ( # ` suc A ) = ( ( # ` A ) + 1 ) )

Proof

Step Hyp Ref Expression
1 nnon
 |-  ( A e. _om -> A e. On )
2 oa1suc
 |-  ( A e. On -> ( A +o 1o ) = suc A )
3 1 2 syl
 |-  ( A e. _om -> ( A +o 1o ) = suc A )
4 3 fveq2d
 |-  ( A e. _om -> ( # ` ( A +o 1o ) ) = ( # ` suc A ) )
5 1onn
 |-  1o e. _om
6 hashnna
 |-  ( ( A e. _om /\ 1o e. _om ) -> ( # ` ( A +o 1o ) ) = ( ( # ` A ) + ( # ` 1o ) ) )
7 5 6 mpan2
 |-  ( A e. _om -> ( # ` ( A +o 1o ) ) = ( ( # ` A ) + ( # ` 1o ) ) )
8 4 7 eqtr3d
 |-  ( A e. _om -> ( # ` suc A ) = ( ( # ` A ) + ( # ` 1o ) ) )
9 hash1
 |-  ( # ` 1o ) = 1
10 9 oveq2i
 |-  ( ( # ` A ) + ( # ` 1o ) ) = ( ( # ` A ) + 1 )
11 8 10 eqtrdi
 |-  ( A e. _om -> ( # ` suc A ) = ( ( # ` A ) + 1 ) )