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 ∈ ω → ( ♯ ‘ suc 𝐴 ) = ( ( ♯ ‘ 𝐴 ) + 1 ) )

Proof

Step Hyp Ref Expression
1 nnon ( 𝐴 ∈ ω → 𝐴 ∈ On )
2 oa1suc ( 𝐴 ∈ On → ( 𝐴 +o 1o ) = suc 𝐴 )
3 1 2 syl ( 𝐴 ∈ ω → ( 𝐴 +o 1o ) = suc 𝐴 )
4 3 fveq2d ( 𝐴 ∈ ω → ( ♯ ‘ ( 𝐴 +o 1o ) ) = ( ♯ ‘ suc 𝐴 ) )
5 1onn 1o ∈ ω
6 hashnna ( ( 𝐴 ∈ ω ∧ 1o ∈ ω ) → ( ♯ ‘ ( 𝐴 +o 1o ) ) = ( ( ♯ ‘ 𝐴 ) + ( ♯ ‘ 1o ) ) )
7 5 6 mpan2 ( 𝐴 ∈ ω → ( ♯ ‘ ( 𝐴 +o 1o ) ) = ( ( ♯ ‘ 𝐴 ) + ( ♯ ‘ 1o ) ) )
8 4 7 eqtr3d ( 𝐴 ∈ ω → ( ♯ ‘ suc 𝐴 ) = ( ( ♯ ‘ 𝐴 ) + ( ♯ ‘ 1o ) ) )
9 hash1 ( ♯ ‘ 1o ) = 1
10 9 oveq2i ( ( ♯ ‘ 𝐴 ) + ( ♯ ‘ 1o ) ) = ( ( ♯ ‘ 𝐴 ) + 1 )
11 8 10 eqtrdi ( 𝐴 ∈ ω → ( ♯ ‘ suc 𝐴 ) = ( ( ♯ ‘ 𝐴 ) + 1 ) )