Metamath Proof Explorer


Theorem erdszelem6

Description: Lemma for erdsze . (Contributed by Mario Carneiro, 22-Jan-2015)

Ref Expression
Hypotheses erdsze.n ( 𝜑𝑁 ∈ ℕ )
erdsze.f ( 𝜑𝐹 : ( 1 ... 𝑁 ) –1-1→ ℝ )
erdszelem.k 𝐾 = ( 𝑥 ∈ ( 1 ... 𝑁 ) ↦ sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } ) , ℝ , < ) )
erdszelem.o 𝑂 Or ℝ
Assertion erdszelem6 ( 𝜑𝐾 : ( 1 ... 𝑁 ) ⟶ ℕ )

Proof

Step Hyp Ref Expression
1 erdsze.n ( 𝜑𝑁 ∈ ℕ )
2 erdsze.f ( 𝜑𝐹 : ( 1 ... 𝑁 ) –1-1→ ℝ )
3 erdszelem.k 𝐾 = ( 𝑥 ∈ ( 1 ... 𝑁 ) ↦ sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } ) , ℝ , < ) )
4 erdszelem.o 𝑂 Or ℝ
5 ltso < Or ℝ
6 5 supex sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } ) , ℝ , < ) ∈ V
7 6 a1i ( ( 𝜑𝑥 ∈ ( 1 ... 𝑁 ) ) → sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } ) , ℝ , < ) ∈ V )
8 3 a1i ( 𝜑𝐾 = ( 𝑥 ∈ ( 1 ... 𝑁 ) ↦ sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } ) , ℝ , < ) ) )
9 eqid { 𝑦 ∈ 𝒫 ( 1 ... 𝑧 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑧𝑦 ) } = { 𝑦 ∈ 𝒫 ( 1 ... 𝑧 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑧𝑦 ) }
10 9 erdszelem2 ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑧 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑧𝑦 ) } ) ∈ Fin ∧ ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑧 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑧𝑦 ) } ) ⊆ ℕ )
11 10 simpri ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑧 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑧𝑦 ) } ) ⊆ ℕ
12 1 2 3 4 erdszelem5 ( ( 𝜑𝑧 ∈ ( 1 ... 𝑁 ) ) → ( 𝐾𝑧 ) ∈ ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑧 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑧𝑦 ) } ) )
13 11 12 sseldi ( ( 𝜑𝑧 ∈ ( 1 ... 𝑁 ) ) → ( 𝐾𝑧 ) ∈ ℕ )
14 7 8 13 fmpt2d ( 𝜑𝐾 : ( 1 ... 𝑁 ) ⟶ ℕ )