Metamath Proof Explorer


Theorem inaex

Description: Assuming the Tarski-Grothendieck axiom, every ordinal is contained in an inaccessible ordinal. (Contributed by Rohan Ridenour, 13-Aug-2023)

Ref Expression
Assertion inaex ( 𝐴 ∈ On → ∃ 𝑥 ∈ Inacc 𝐴𝑥 )

Proof

Step Hyp Ref Expression
1 inawina ( 𝑥 ∈ Inacc → 𝑥 ∈ Inaccw )
2 winaon ( 𝑥 ∈ Inaccw𝑥 ∈ On )
3 1 2 syl ( 𝑥 ∈ Inacc → 𝑥 ∈ On )
4 3 ssriv Inacc ⊆ On
5 onmindif ( ( Inacc ⊆ On ∧ 𝐴 ∈ On ) → 𝐴 ( Inacc ∖ suc 𝐴 ) )
6 4 5 mpan ( 𝐴 ∈ On → 𝐴 ( Inacc ∖ suc 𝐴 ) )
7 6 adantr ( ( 𝐴 ∈ On ∧ 𝑥 = ( Inacc ∖ suc 𝐴 ) ) → 𝐴 ( Inacc ∖ suc 𝐴 ) )
8 simpr ( ( 𝐴 ∈ On ∧ 𝑥 = ( Inacc ∖ suc 𝐴 ) ) → 𝑥 = ( Inacc ∖ suc 𝐴 ) )
9 7 8 eleqtrrd ( ( 𝐴 ∈ On ∧ 𝑥 = ( Inacc ∖ suc 𝐴 ) ) → 𝐴𝑥 )
10 difss ( Inacc ∖ suc 𝐴 ) ⊆ Inacc
11 10 4 sstri ( Inacc ∖ suc 𝐴 ) ⊆ On
12 inaprc Inacc ∉ V
13 12 neli ¬ Inacc ∈ V
14 ssdif0 ( Inacc ⊆ suc 𝐴 ↔ ( Inacc ∖ suc 𝐴 ) = ∅ )
15 sucexg ( 𝐴 ∈ On → suc 𝐴 ∈ V )
16 ssexg ( ( Inacc ⊆ suc 𝐴 ∧ suc 𝐴 ∈ V ) → Inacc ∈ V )
17 16 expcom ( suc 𝐴 ∈ V → ( Inacc ⊆ suc 𝐴 → Inacc ∈ V ) )
18 15 17 syl ( 𝐴 ∈ On → ( Inacc ⊆ suc 𝐴 → Inacc ∈ V ) )
19 14 18 syl5bir ( 𝐴 ∈ On → ( ( Inacc ∖ suc 𝐴 ) = ∅ → Inacc ∈ V ) )
20 13 19 mtoi ( 𝐴 ∈ On → ¬ ( Inacc ∖ suc 𝐴 ) = ∅ )
21 20 neqned ( 𝐴 ∈ On → ( Inacc ∖ suc 𝐴 ) ≠ ∅ )
22 onint ( ( ( Inacc ∖ suc 𝐴 ) ⊆ On ∧ ( Inacc ∖ suc 𝐴 ) ≠ ∅ ) → ( Inacc ∖ suc 𝐴 ) ∈ ( Inacc ∖ suc 𝐴 ) )
23 11 21 22 sylancr ( 𝐴 ∈ On → ( Inacc ∖ suc 𝐴 ) ∈ ( Inacc ∖ suc 𝐴 ) )
24 23 eldifad ( 𝐴 ∈ On → ( Inacc ∖ suc 𝐴 ) ∈ Inacc )
25 9 24 rspcime ( 𝐴 ∈ On → ∃ 𝑥 ∈ Inacc 𝐴𝑥 )