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 1 𝑜 ∈ ran ⁡ card

Proof

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