Metamath Proof Explorer


Theorem sucomisnotcard

Description: _om +o 1o is not a cardinal number. (Contributed by RP, 1-Oct-2023)

Ref Expression
Assertion sucomisnotcard ¬ ( ω +o 1o ) ∈ ran card

Proof

Step Hyp Ref Expression
1 omelon ω ∈ On
2 sucidg ( ω ∈ On → ω ∈ suc ω )
3 1 2 ax-mp ω ∈ suc ω
4 omensuc ω ≈ suc ω
5 breq1 ( 𝑥 = ω → ( 𝑥 ≈ suc ω ↔ ω ≈ suc ω ) )
6 5 rspcev ( ( ω ∈ suc ω ∧ ω ≈ suc ω ) → ∃ 𝑥 ∈ suc ω 𝑥 ≈ suc ω )
7 3 4 6 mp2an 𝑥 ∈ suc ω 𝑥 ≈ suc ω
8 dfrex2 ( ∃ 𝑥 ∈ suc ω 𝑥 ≈ suc ω ↔ ¬ ∀ 𝑥 ∈ suc ω ¬ 𝑥 ≈ suc ω )
9 7 8 mpbi ¬ ∀ 𝑥 ∈ suc ω ¬ 𝑥 ≈ suc ω
10 9 intnan ¬ ( suc ω ∈ On ∧ ∀ 𝑥 ∈ suc ω ¬ 𝑥 ≈ suc ω )
11 oa1suc ( ω ∈ On → ( ω +o 1o ) = suc ω )
12 1 11 ax-mp ( ω +o 1o ) = suc ω
13 12 eleq1i ( ( ω +o 1o ) ∈ ran card ↔ suc ω ∈ ran card )
14 elrncard ( suc ω ∈ ran card ↔ ( suc ω ∈ On ∧ ∀ 𝑥 ∈ suc ω ¬ 𝑥 ≈ suc ω ) )
15 13 14 sylbb ( ( ω +o 1o ) ∈ ran card → ( suc ω ∈ On ∧ ∀ 𝑥 ∈ suc ω ¬ 𝑥 ≈ suc ω ) )
16 10 15 mto ¬ ( ω +o 1o ) ∈ ran card