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. lexicographic path order with precedence:
    
      precedence:
      
        del# > = > . > u > and > v > nil > true > false > f#
      
      argument filter:
    
        pi(del#) = 1
        pi(.) = 2
        pi(f#) = 4
        pi(=) = [1]
        pi(false) = []
        pi(true) = []
        pi(nil) = []
        pi(u) = []
        pi(v) = []
        pi(and) = 2
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        del#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,1,1,1)) x1
        ._A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (2,1,1,1)
        f#_A(x1,x2,x3,x4) = x4 + (3,0,0,4)
        =_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,0,0)
        false_A() = (1,1,1,1)
        true_A() = (1,1,1,1)
        nil_A() = (0,0,0,0)
        u_A() = (0,0,0,0)
        v_A() = (1,1,1,1)
        and_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x2
    

The next rules are strictly ordered:

  p1, p2, p3

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