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I have a system of inequalities and I am wondering wether or not there exists a solution.

With these definitions

a1 = (piS (1 - y))/(1 - x + (x - y) piS)
a2 = (piS (1 - q y))/(1 - q x + q piS (x - y))

a3 = (piS y)/ (x (1 - piS) + y piS)

P1 = (b (q y - 1) + k a2)/(k - b q (x - y))

P2 = (k a3 - b q y)/(b q (x - y) + k)

P3 = (k a2 - b)/k

P4 = (b (q y - 1) + k (a2 - a3))/(b q (y - x))

the question is whether or not there exist a solution of

P4 > P3 && P4 > 0 && P4 < a3 && P3 < P2 && P1 < a2 && P1 > 0 && 
q > 0 && q < 1 && x > 1/2 && x < 1 && y < 1/2 && y > 0 && 1 > pi && 
pi > 0 && a3 < a2 &&  a2 < a1 && piS > 0 && piS < 1 &&  k > 0 && 
b > 0 &&  a2 > 0 && a2 < 1 && a3 > 0 && a1 < 1

I have tried Fullsimplify and it does not output false. However, FindInstance applied to the result does not give an example (after 2 hours it is still running).

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The answer is affirmative. We put q, x, piS on our own Then FindInstance works quickly:

q = 0.5; piS = 0.5; x = 0.75;
FindInstance[P4 > P3 && P4 > 0 && P4 < a3 && P3 < P2 && P1 < a2 && P1 > 0 && 
  q > 0 && q < 1 && x > 1/2 && x < 1 && y < 1/2 && y > 0 && 1 > piS && 
  piS > 0 && a3 < a2 && a2 < a1 && piS > 0 && piS < 1 && k > 0 && 
  b > 0 && a2 > 0 && a2 < 1 && a3 > 0 && a1 < 1, {y, b, k}, Reals]

{{y->0.25,b->0.40625,k->1.}}

Notice. Few typos in your code are corrected - piS instead of pi .

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