YES

We show the termination of the TRS R:

  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()))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: del#(.(x,.(y,z))) -> f#(=(x,y),x,y,z)
p2: del#(.(x,.(y,z))) -> =#(x,y)
p3: f#(true(),x,y,z) -> del#(.(y,z))
p4: f#(false(),x,y,z) -> del#(.(y,z))
p5: =#(.(x,y),.(u(),v())) -> =#(x,u())
p6: =#(.(x,y),.(u(),v())) -> =#(y,v())

and R consists of:

r1: del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
r2: f(true(),x,y,z) -> del(.(y,z))
r3: f(false(),x,y,z) -> .(x,del(.(y,z)))
r4: =(nil(),nil()) -> true()
r5: =(.(x,y),nil()) -> false()
r6: =(nil(),.(y,z)) -> false()
r7: =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: del#(.(x,.(y,z))) -> f#(=(x,y),x,y,z)
p2: f#(false(),x,y,z) -> del#(.(y,z))
p3: f#(true(),x,y,z) -> del#(.(y,z))

and R consists of:

r1: del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
r2: f(true(),x,y,z) -> del(.(y,z))
r3: f(false(),x,y,z) -> .(x,del(.(y,z)))
r4: =(nil(),nil()) -> true()
r5: =(.(x,y),nil()) -> false()
r6: =(nil(),.(y,z)) -> false()
r7: =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))

The set of usable rules consists of

  r4, r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            del#_A(x1) = x1 + (1,3)
            ._A(x1,x2) = x1 + x2 + (4,3)
            f#_A(x1,x2,x3,x4) = x3 + x4 + (6,7)
            =_A(x1,x2) = (2,10)
            false_A() = (1,2)
            true_A() = (1,2)
            nil_A() = (2,1)
            u_A() = (3,0)
            v_A() = (1,0)
            and_A(x1,x2) = (0,0)
    
      precedence:
      
        false > f# > del# > . = = = u = v > true = nil > and
      
      partial status:
    
        pi(del#) = []
        pi(.) = [1, 2]
        pi(f#) = [4]
        pi(=) = []
        pi(false) = []
        pi(true) = []
        pi(nil) = []
        pi(u) = []
        pi(v) = []
        pi(and) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            del#_A(x1) = (0,0)
            ._A(x1,x2) = x2
            f#_A(x1,x2,x3,x4) = ((1,0),(1,1)) x4 + (1,1)
            =_A(x1,x2) = (3,0)
            false_A() = (0,0)
            true_A() = (1,1)
            nil_A() = (0,0)
            u_A() = (1,0)
            v_A() = (1,0)
            and_A(x1,x2) = (2,0)
    
      precedence:
      
        u > = = and > v > del# = . = f# = false = true = nil
      
      partial status:
    
        pi(del#) = []
        pi(.) = []
        pi(f#) = [4]
        pi(=) = []
        pi(false) = []
        pi(true) = []
        pi(nil) = []
        pi(u) = []
        pi(v) = []
        pi(and) = []
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.