Metamath Proof Explorer


Theorem rankkardu

Description: An upper bound on the rank of a kard cardinal. (Contributed by BTernaryTau, 4-Jul-2026)

Ref Expression
Assertion rankkardu ( rank ‘ ( kard ‘ 𝐴 ) ) ⊆ suc ( rank ‘ 𝐴 )

Proof

Step Hyp Ref Expression
1 kardval ⊢ ( kard ‘ 𝐴 ) = Scott { 𝑥 ∣ 𝑥 ≈ 𝐴 }
2 1 fveq2i ⊢ ( rank ‘ ( kard ‘ 𝐴 ) ) = ( rank ‘ Scott { 𝑥 ∣ 𝑥 ≈ 𝐴 } )
3 enrefg ⊢ ( 𝐴 ∈ V → 𝐴 ≈ 𝐴 )
4 breq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 ≈ 𝐴 ↔ 𝐴 ≈ 𝐴 ) )
5 4 elabg ⊢ ( 𝐴 ∈ V → ( 𝐴 ∈ { 𝑥 ∣ 𝑥 ≈ 𝐴 } ↔ 𝐴 ≈ 𝐴 ) )
6 3 5 mpbird ⊢ ( 𝐴 ∈ V → 𝐴 ∈ { 𝑥 ∣ 𝑥 ≈ 𝐴 } )
7 rankscottu ⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝑥 ≈ 𝐴 } → ( rank ‘ Scott { 𝑥 ∣ 𝑥 ≈ 𝐴 } ) ⊆ suc ( rank ‘ 𝐴 ) )
8 6 7 syl ⊢ ( 𝐴 ∈ V → ( rank ‘ Scott { 𝑥 ∣ 𝑥 ≈ 𝐴 } ) ⊆ suc ( rank ‘ 𝐴 ) )
9 2 8 eqsstrid ⊢ ( 𝐴 ∈ V → ( rank ‘ ( kard ‘ 𝐴 ) ) ⊆ suc ( rank ‘ 𝐴 ) )
10 kardeq0 ⊢ ( ( kard ‘ 𝐴 ) = ∅ ↔ ¬ 𝐴 ∈ V )
11 fvex ⊢ ( kard ‘ 𝐴 ) ∈ V
12 11 rankeq0 ⊢ ( ( kard ‘ 𝐴 ) = ∅ ↔ ( rank ‘ ( kard ‘ 𝐴 ) ) = ∅ )
13 0ss ⊢ ∅ ⊆ suc ( rank ‘ 𝐴 )
14 sseq1 ⊢ ( ( rank ‘ ( kard ‘ 𝐴 ) ) = ∅ → ( ( rank ‘ ( kard ‘ 𝐴 ) ) ⊆ suc ( rank ‘ 𝐴 ) ↔ ∅ ⊆ suc ( rank ‘ 𝐴 ) ) )
15 13 14 mpbiri ⊢ ( ( rank ‘ ( kard ‘ 𝐴 ) ) = ∅ → ( rank ‘ ( kard ‘ 𝐴 ) ) ⊆ suc ( rank ‘ 𝐴 ) )
16 12 15 sylbi ⊢ ( ( kard ‘ 𝐴 ) = ∅ → ( rank ‘ ( kard ‘ 𝐴 ) ) ⊆ suc ( rank ‘ 𝐴 ) )
17 10 16 sylbir ⊢ ( ¬ 𝐴 ∈ V → ( rank ‘ ( kard ‘ 𝐴 ) ) ⊆ suc ( rank ‘ 𝐴 ) )
18 9 17 pm2.61i ⊢ ( rank ‘ ( kard ‘ 𝐴 ) ) ⊆ suc ( rank ‘ 𝐴 )