Metamath Proof Explorer


Theorem ballotlemfval0

Description: ( FC ) always starts counting at 0 . (Contributed by Thierry Arnoux, 25-Nov-2016)

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 = c ∈ O ⟼ i ∈ ℤ ⟼ 1 … i ∩ c − 1 … i ∖ c
Assertion ballotlemfval0 ⊢ C ∈ O → F ⁡ C ⁡ 0 = 0

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 = c ∈ O ⟼ i ∈ ℤ ⟼ 1 … i ∩ c − 1 … i ∖ c
6 id ⊢ C ∈ O → C ∈ O
7 0zd ⊢ C ∈ O → 0 ∈ ℤ
8 1 2 3 4 5 6 7 ballotlemfval ⊢ C ∈ O → F ⁡ C ⁡ 0 = 1 … 0 ∩ C − 1 … 0 ∖ C
9 fz10 ⊢ 1 … 0 = ∅
10 9 ineq1i ⊢ 1 … 0 ∩ C = ∅ ∩ C
11 incom ⊢ C ∩ ∅ = ∅ ∩ C
12 in0 ⊢ C ∩ ∅ = ∅
13 10 11 12 3eqtr2i ⊢ 1 … 0 ∩ C = ∅
14 13 fveq2i ⊢ 1 … 0 ∩ C = ∅
15 hash0 ⊢ ∅ = 0
16 14 15 eqtri ⊢ 1 … 0 ∩ C = 0
17 9 difeq1i ⊢ 1 … 0 ∖ C = ∅ ∖ C
18 0dif ⊢ ∅ ∖ C = ∅
19 17 18 eqtri ⊢ 1 … 0 ∖ C = ∅
20 19 fveq2i ⊢ 1 … 0 ∖ C = ∅
21 20 15 eqtri ⊢ 1 … 0 ∖ C = 0
22 16 21 oveq12i ⊢ 1 … 0 ∩ C − 1 … 0 ∖ C = 0 − 0
23 0m0e0 ⊢ 0 − 0 = 0
24 22 23 eqtri ⊢ 1 … 0 ∩ C − 1 … 0 ∖ C = 0
25 8 24 eqtrdi ⊢ C ∈ O → F ⁡ C ⁡ 0 = 0