Metamath Proof Explorer


Theorem erdszelem3

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 , < ) )
Assertion erdszelem3 ( 𝐴 ∈ ( 1 ... 𝑁 ) → ( 𝐾𝐴 ) = sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝐴 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝐴𝑦 ) } ) , ℝ , < ) )

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 oveq2 ( 𝑥 = 𝐴 → ( 1 ... 𝑥 ) = ( 1 ... 𝐴 ) )
5 4 pweqd ( 𝑥 = 𝐴 → 𝒫 ( 1 ... 𝑥 ) = 𝒫 ( 1 ... 𝐴 ) )
6 eleq1 ( 𝑥 = 𝐴 → ( 𝑥𝑦𝐴𝑦 ) )
7 6 anbi2d ( 𝑥 = 𝐴 → ( ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) ↔ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝐴𝑦 ) ) )
8 5 7 rabeqbidv ( 𝑥 = 𝐴 → { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } = { 𝑦 ∈ 𝒫 ( 1 ... 𝐴 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝐴𝑦 ) } )
9 8 imaeq2d ( 𝑥 = 𝐴 → ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } ) = ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝐴 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝐴𝑦 ) } ) )
10 9 supeq1d ( 𝑥 = 𝐴 → sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝑥 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝑥𝑦 ) } ) , ℝ , < ) = sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝐴 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝐴𝑦 ) } ) , ℝ , < ) )
11 ltso < Or ℝ
12 11 supex sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝐴 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝐴𝑦 ) } ) , ℝ , < ) ∈ V
13 10 3 12 fvmpt ( 𝐴 ∈ ( 1 ... 𝑁 ) → ( 𝐾𝐴 ) = sup ( ( ♯ “ { 𝑦 ∈ 𝒫 ( 1 ... 𝐴 ) ∣ ( ( 𝐹𝑦 ) Isom < , 𝑂 ( 𝑦 , ( 𝐹𝑦 ) ) ∧ 𝐴𝑦 ) } ) , ℝ , < ) )