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:
  
  del#(.(x, .(y, z))) -> f#(=(x, y), x, y, z)
  f#(true, x, y, z) -> del#(.(y, z))
  f#(false, x, y, z) -> del#(.(y, z))
  rules:
  
  =(nil, nil) -> true
  =(.(x, y), nil) -> false
  =(nil, .(y, z)) -> false
  =(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))
  del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
  f(true, x, y, z) -> del(.(y, z))
  f(false, x, y, z) -> .(x, del(.(y, z)))
  
   the pairs 
  del#(.(x, .(y, z))) -> f#(=(x, y), x, y, z)
  f#(true, x, y, z) -> del#(.(y, z))
  f#(false, x, y, z) -> del#(.(y, z))
  
  could not apply the generic root reduction pair processor with the following
  SCNP-version with mu = MS and the level mapping defined by 
  pi(f#) = [(epsilon,0),(1,0),(4,1)]
  pi(del#) = [(epsilon,0),(1,0)]
  Argument Filter: 
  pi(f#/4) = [4]
  pi(true/0) = []
  pi(del#/1) = []
  pi(./2) = [2]
  pi(=/2) = []
  pi(false/0) = []
  pi(nil/0) = []
  pi(u/0) = []
  pi(v/0) = []
  pi(and/2) = 1
  
  RPO with the following precedence
  precedence(f#[4]) = 0
  precedence(true[0]) = 1
  precedence(del#[1]) = 1
  precedence(.[2]) = 0
  precedence(=[2]) = 1
  precedence(false[0]) = 1
  precedence(nil[0]) = 2
  precedence(u[0]) = 3
  precedence(v[0]) = 4
  
  precedence(_) = 0
  and the following status
  status(f#[4]) = lex
  status(true[0]) = lex
  status(del#[1]) = lex
  status(.[2]) = lex
  status(=[2]) = lex
  status(false[0]) = lex
  status(nil[0]) = lex
  status(u[0]) = lex
  status(v[0]) = lex
  
  status(_) = lex
  
  problem when orienting DPs
  cannot orient pair del#(.(x, .(y, z))) -> f#(=(x, y), x, y, z) strictly:
  [(del#(.(x, .(y, z))),0),(.(x, .(y, z)),0)] >mu [(f#(=(x, y), x, y, z),0),(=(x, y),0),(z,1)] could not be ensured