Metamath Proof Explorer


Theorem ballotlemrval

Description: Value of R . (Contributed by Thierry Arnoux, 14-Apr-2017)

Ref Expression
Hypotheses ballotth.m 𝑀 ∈ ℕ
ballotth.n 𝑁 ∈ ℕ
ballotth.o 𝑂 = { 𝑐 ∈ 𝒫 ( 1 ... ( 𝑀 + 𝑁 ) ) ∣ ( ♯ ‘ 𝑐 ) = 𝑀 }
ballotth.p 𝑃 = ( 𝑥 ∈ 𝒫 𝑂 ↦ ( ( ♯ ‘ 𝑥 ) / ( ♯ ‘ 𝑂 ) ) )
ballotth.f 𝐹 = ( 𝑐𝑂 ↦ ( 𝑖 ∈ ℤ ↦ ( ( ♯ ‘ ( ( 1 ... 𝑖 ) ∩ 𝑐 ) ) − ( ♯ ‘ ( ( 1 ... 𝑖 ) ∖ 𝑐 ) ) ) ) )
ballotth.e 𝐸 = { 𝑐𝑂 ∣ ∀ 𝑖 ∈ ( 1 ... ( 𝑀 + 𝑁 ) ) 0 < ( ( 𝐹𝑐 ) ‘ 𝑖 ) }
ballotth.mgtn 𝑁 < 𝑀
ballotth.i 𝐼 = ( 𝑐 ∈ ( 𝑂𝐸 ) ↦ inf ( { 𝑘 ∈ ( 1 ... ( 𝑀 + 𝑁 ) ) ∣ ( ( 𝐹𝑐 ) ‘ 𝑘 ) = 0 } , ℝ , < ) )
ballotth.s 𝑆 = ( 𝑐 ∈ ( 𝑂𝐸 ) ↦ ( 𝑖 ∈ ( 1 ... ( 𝑀 + 𝑁 ) ) ↦ if ( 𝑖 ≤ ( 𝐼𝑐 ) , ( ( ( 𝐼𝑐 ) + 1 ) − 𝑖 ) , 𝑖 ) ) )
ballotth.r 𝑅 = ( 𝑐 ∈ ( 𝑂𝐸 ) ↦ ( ( 𝑆𝑐 ) “ 𝑐 ) )
Assertion ballotlemrval ( 𝐶 ∈ ( 𝑂𝐸 ) → ( 𝑅𝐶 ) = ( ( 𝑆𝐶 ) “ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 ballotth.m 𝑀 ∈ ℕ
2 ballotth.n 𝑁 ∈ ℕ
3 ballotth.o 𝑂 = { 𝑐 ∈ 𝒫 ( 1 ... ( 𝑀 + 𝑁 ) ) ∣ ( ♯ ‘ 𝑐 ) = 𝑀 }
4 ballotth.p 𝑃 = ( 𝑥 ∈ 𝒫 𝑂 ↦ ( ( ♯ ‘ 𝑥 ) / ( ♯ ‘ 𝑂 ) ) )
5 ballotth.f 𝐹 = ( 𝑐𝑂 ↦ ( 𝑖 ∈ ℤ ↦ ( ( ♯ ‘ ( ( 1 ... 𝑖 ) ∩ 𝑐 ) ) − ( ♯ ‘ ( ( 1 ... 𝑖 ) ∖ 𝑐 ) ) ) ) )
6 ballotth.e 𝐸 = { 𝑐𝑂 ∣ ∀ 𝑖 ∈ ( 1 ... ( 𝑀 + 𝑁 ) ) 0 < ( ( 𝐹𝑐 ) ‘ 𝑖 ) }
7 ballotth.mgtn 𝑁 < 𝑀
8 ballotth.i 𝐼 = ( 𝑐 ∈ ( 𝑂𝐸 ) ↦ inf ( { 𝑘 ∈ ( 1 ... ( 𝑀 + 𝑁 ) ) ∣ ( ( 𝐹𝑐 ) ‘ 𝑘 ) = 0 } , ℝ , < ) )
9 ballotth.s 𝑆 = ( 𝑐 ∈ ( 𝑂𝐸 ) ↦ ( 𝑖 ∈ ( 1 ... ( 𝑀 + 𝑁 ) ) ↦ if ( 𝑖 ≤ ( 𝐼𝑐 ) , ( ( ( 𝐼𝑐 ) + 1 ) − 𝑖 ) , 𝑖 ) ) )
10 ballotth.r 𝑅 = ( 𝑐 ∈ ( 𝑂𝐸 ) ↦ ( ( 𝑆𝑐 ) “ 𝑐 ) )
11 fveq2 ( 𝑑 = 𝐶 → ( 𝑆𝑑 ) = ( 𝑆𝐶 ) )
12 id ( 𝑑 = 𝐶𝑑 = 𝐶 )
13 11 12 imaeq12d ( 𝑑 = 𝐶 → ( ( 𝑆𝑑 ) “ 𝑑 ) = ( ( 𝑆𝐶 ) “ 𝐶 ) )
14 fveq2 ( 𝑐 = 𝑑 → ( 𝑆𝑐 ) = ( 𝑆𝑑 ) )
15 id ( 𝑐 = 𝑑𝑐 = 𝑑 )
16 14 15 imaeq12d ( 𝑐 = 𝑑 → ( ( 𝑆𝑐 ) “ 𝑐 ) = ( ( 𝑆𝑑 ) “ 𝑑 ) )
17 16 cbvmptv ( 𝑐 ∈ ( 𝑂𝐸 ) ↦ ( ( 𝑆𝑐 ) “ 𝑐 ) ) = ( 𝑑 ∈ ( 𝑂𝐸 ) ↦ ( ( 𝑆𝑑 ) “ 𝑑 ) )
18 10 17 eqtri 𝑅 = ( 𝑑 ∈ ( 𝑂𝐸 ) ↦ ( ( 𝑆𝑑 ) “ 𝑑 ) )
19 fvex ( 𝑆𝐶 ) ∈ V
20 imaexg ( ( 𝑆𝐶 ) ∈ V → ( ( 𝑆𝐶 ) “ 𝐶 ) ∈ V )
21 19 20 ax-mp ( ( 𝑆𝐶 ) “ 𝐶 ) ∈ V
22 13 18 21 fvmpt ( 𝐶 ∈ ( 𝑂𝐸 ) → ( 𝑅𝐶 ) = ( ( 𝑆𝐶 ) “ 𝐶 ) )