Metamath Proof Explorer


Theorem ballotlemieq

Description: If two countings share the same first tie, they also have the same swap function. (Contributed by Thierry Arnoux, 18-Apr-2017)

Ref Expression
Hypotheses ballotth.m M
ballotth.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 1 i c 1 i c
ballotth.e E = c O | i 1 M + N 0 < F c i
ballotth.mgtn N < M
ballotth.i I = c O E sup k 1 M + N | F c k = 0 <
ballotth.s S = c O E i 1 M + N if i I c I c + 1 - i i
Assertion ballotlemieq C O E D O E I C = I D S C = S D

Proof

Step Hyp Ref Expression
1 ballotth.m M
2 ballotth.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 1 i c 1 i c
6 ballotth.e E = c O | i 1 M + N 0 < F c i
7 ballotth.mgtn N < M
8 ballotth.i I = c O E sup k 1 M + N | F c k = 0 <
9 ballotth.s S = c O E i 1 M + N if i I c I c + 1 - i i
10 simpl I C = I D i 1 M + N I C = I D
11 10 breq2d I C = I D i 1 M + N i I C i I D
12 10 oveq1d I C = I D i 1 M + N I C + 1 = I D + 1
13 12 oveq1d I C = I D i 1 M + N I C + 1 - i = I D + 1 - i
14 11 13 ifbieq1d I C = I D i 1 M + N if i I C I C + 1 - i i = if i I D I D + 1 - i i
15 14 mpteq2dva I C = I D i 1 M + N if i I C I C + 1 - i i = i 1 M + N if i I D I D + 1 - i i
16 15 3ad2ant3 C O E D O E I C = I D i 1 M + N if i I C I C + 1 - i i = i 1 M + N if i I D I D + 1 - i i
17 1 2 3 4 5 6 7 8 9 ballotlemsval C O E S C = i 1 M + N if i I C I C + 1 - i i
18 17 3ad2ant1 C O E D O E I C = I D S C = i 1 M + N if i I C I C + 1 - i i
19 1 2 3 4 5 6 7 8 9 ballotlemsval D O E S D = i 1 M + N if i I D I D + 1 - i i
20 19 3ad2ant2 C O E D O E I C = I D S D = i 1 M + N if i I D I D + 1 - i i
21 16 18 20 3eqtr4d C O E D O E I C = I D S C = S D