ceta_eq: termination proof not accepted
1: error below switch to dependency pairs
1.1: error below the dependency graph processor
 1.1.1: error when applying the reduction pair processor with usable rules to remove from the DP problem
  pairs:
  
  selsort#(cons(N, L)) -> ifselsort#(eq(N, min(cons(N, L))), cons(N, L))
  ifselsort#(true, cons(N, L)) -> selsort#(L)
  ifselsort#(false, cons(N, L)) -> selsort#(replace(min(cons(N, L)), N, L))
  rules:
  
  eq(0, 0) -> true
  eq(0, s(Y)) -> false
  eq(s(X), 0) -> false
  eq(s(X), s(Y)) -> eq(X, Y)
  ifmin(true, cons(N, cons(M, L))) -> min(cons(N, L))
  ifmin(false, cons(N, cons(M, L))) -> min(cons(M, L))
  ifrepl(true, N, M, cons(K, L)) -> cons(M, L)
  ifrepl(false, N, M, cons(K, L)) -> cons(K, replace(N, M, L))
  ifselsort(true, cons(N, L)) -> cons(N, selsort(L))
  ifselsort(false, cons(N, L)) -> cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))
  le(0, Y) -> true
  le(s(X), 0) -> false
  le(s(X), s(Y)) -> le(X, Y)
  min(cons(0, nil)) -> 0
  min(cons(s(N), nil)) -> s(N)
  min(cons(N, cons(M, L))) -> ifmin(le(N, M), cons(N, cons(M, L)))
  replace(N, M, nil) -> nil
  replace(N, M, cons(K, L)) -> ifrepl(eq(N, K), N, M, cons(K, L))
  selsort(nil) -> nil
  selsort(cons(N, L)) -> ifselsort(eq(N, min(cons(N, L))), cons(N, L))
  
   the pairs 
  ifselsort#(true, cons(N, L)) -> selsort#(L)
  
  could not apply the generic root reduction pair processor with the following
  SCNP-version with mu = MAX and the level mapping defined by 
  pi(ifselsort#) = [(epsilon,0),(1,0)]
  pi(selsort#) = [(1,0)]
  Argument Filter: 
  pi(ifselsort#/2) = 2
  pi(true/0) = []
  pi(cons/2) = [2]
  pi(selsort#/1) = []
  pi(eq/2) = []
  pi(min/1) = []
  pi(false/0) = []
  pi(replace/3) = [3]
  pi(0/0) = []
  pi(nil/0) = []
  pi(s/1) = 1
  pi(ifmin/2) = []
  pi(le/2) = [1,2]
  pi(ifrepl/4) = [4]
  
  RPO with the following precedence
  precedence(true[0]) = 1
  precedence(cons[2]) = 5
  precedence(selsort#[1]) = 5
  precedence(eq[2]) = 5
  precedence(min[1]) = 6
  precedence(false[0]) = 2
  precedence(replace[3]) = 5
  precedence(0[0]) = 3
  precedence(nil[0]) = 5
  precedence(ifmin[2]) = 4
  precedence(le[2]) = 0
  precedence(ifrepl[4]) = 5
  
  precedence(_) = 0
  and the following status
  status(true[0]) = mul
  status(cons[2]) = mul
  status(selsort#[1]) = mul
  status(eq[2]) = mul
  status(min[1]) = mul
  status(false[0]) = mul
  status(replace[3]) = mul
  status(0[0]) = mul
  status(nil[0]) = mul
  status(ifmin[2]) = mul
  status(le[2]) = mul
  status(ifrepl[4]) = mul
  
  status(_) = lex
  
  problem when orienting (usable) rules
  could not orient ifrepl(false, N, M, cons(K, L)) >= cons(K, replace(N, M, L))
  pi( ifrepl(false, N, M, cons(K, L)) ) = ifrepl(cons(L))
  pi( cons(K, replace(N, M, L)) ) = cons(replace(L))
  could not orient ifrepl(cons(L)) >=RPO cons(replace(L))