YES

Problem:
 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)))
 =(nil(),nil()) -> true()
 =(.(x,y),nil()) -> false()
 =(nil(),.(y,z)) -> false()
 =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))

Proof:
 DP Processor:
  DPs:
   del#(.(x,.(y,z))) -> =#(x,y)
   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))
   =#(.(x,y),.(u(),v())) -> =#(y,v())
   =#(.(x,y),.(u(),v())) -> =#(x,u())
  TRS:
   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)))
   =(nil(),nil()) -> true()
   =(.(x,y),nil()) -> false()
   =(nil(),.(y,z)) -> false()
   =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f#](x0, x1, x2, x3) = 2x1 + 3x2 + 2x3 + 4,
    
    [=#](x0, x1) = 4x1 + 4,
    
    [del#](x0) = x0 + 2,
    
    [and](x0, x1) = 2,
    
    [v] = 0,
    
    [u] = 0,
    
    [nil] = 0,
    
    [false] = 0,
    
    [true] = 0,
    
    [f](x0, x1, x2, x3) = 2x1 + 4x2 + 4x3 + 3,
    
    [=](x0, x1) = 6,
    
    [del](x0) = x0,
    
    [.](x0, x1) = 2x0 + 2x1 + 1
   orientation:
    del#(.(x,.(y,z))) = 2x + 4y + 4z + 5 >= 4y + 4 = =#(x,y)
    
    del#(.(x,.(y,z))) = 2x + 4y + 4z + 5 >= 2x + 3y + 2z + 4 = f#(=(x,y),x,y,z)
    
    f#(true(),x,y,z) = 2x + 3y + 2z + 4 >= 2y + 2z + 3 = del#(.(y,z))
    
    f#(false(),x,y,z) = 2x + 3y + 2z + 4 >= 2y + 2z + 3 = del#(.(y,z))
    
    =#(.(x,y),.(u(),v())) = 8 >= 4 = =#(y,v())
    
    =#(.(x,y),.(u(),v())) = 8 >= 4 = =#(x,u())
    
    del(.(x,.(y,z))) = 2x + 4y + 4z + 3 >= 2x + 4y + 4z + 3 = f(=(x,y),x,y,z)
    
    f(true(),x,y,z) = 2x + 4y + 4z + 3 >= 2y + 2z + 1 = del(.(y,z))
    
    f(false(),x,y,z) = 2x + 4y + 4z + 3 >= 2x + 4y + 4z + 3 = .(x,del(.(y,z)))
    
    =(nil(),nil()) = 6 >= 0 = true()
    
    =(.(x,y),nil()) = 6 >= 0 = false()
    
    =(nil(),.(y,z)) = 6 >= 0 = false()
    
    =(.(x,y),.(u(),v())) = 6 >= 2 = and(=(x,u()),=(y,v()))
   problem:
    DPs:
     
    TRS:
     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)))
     =(nil(),nil()) -> true()
     =(.(x,y),nil()) -> false()
     =(nil(),.(y,z)) -> false()
     =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))
   Qed