Metamath Proof Explorer


Theorem nna1iscard

Description: For any natural number, the add one operation is results in a cardinal number. (Contributed by RP, 1-Oct-2023)

Ref Expression
Assertion nna1iscard ( 𝑁 ∈ ω → ( 𝑁 +o 1o ) ∈ ran card )

Proof

Step Hyp Ref Expression
1 nnon ⊢ ( 𝑁 ∈ ω → 𝑁 ∈ On )
2 oa1suc ⊢ ( 𝑁 ∈ On → ( 𝑁 +o 1o ) = suc 𝑁 )
3 1 2 syl ⊢ ( 𝑁 ∈ ω → ( 𝑁 +o 1o ) = suc 𝑁 )
4 peano2 ⊢ ( 𝑁 ∈ ω → suc 𝑁 ∈ ω )
5 3 4 jca ⊢ ( 𝑁 ∈ ω → ( ( 𝑁 +o 1o ) = suc 𝑁 ∧ suc 𝑁 ∈ ω ) )
6 simpl ⊢ ( ( ( 𝑁 +o 1o ) = suc 𝑁 ∧ suc 𝑁 ∈ ω ) → ( 𝑁 +o 1o ) = suc 𝑁 )
7 simpr ⊢ ( ( ( 𝑁 +o 1o ) = suc 𝑁 ∧ suc 𝑁 ∈ ω ) → suc 𝑁 ∈ ω )
8 6 7 eqeltrd ⊢ ( ( ( 𝑁 +o 1o ) = suc 𝑁 ∧ suc 𝑁 ∈ ω ) → ( 𝑁 +o 1o ) ∈ ω )
9 omssrncard ⊢ ω ⊆ ran card
10 9 sseli ⊢ ( ( 𝑁 +o 1o ) ∈ ω → ( 𝑁 +o 1o ) ∈ ran card )
11 5 8 10 3syl ⊢ ( 𝑁 ∈ ω → ( 𝑁 +o 1o ) ∈ ran card )