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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          del#_A(x1) = ((1,0),(0,0)) x1 + (2,3)
          ._A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(1,0)) x2 + (1,4)
          f#_A(x1,x2,x3,x4) = ((0,1),(0,0)) x3 + ((1,1),(0,0)) x4 + (7,3)
          =_A(x1,x2) = ((0,0),(1,0)) x1 + (0,13)
          false_A() = (0,5)
          true_A() = (0,2)
          nil_A() = (1,1)
          u_A() = (2,5)
          v_A() = (3,6)
          and_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (0,13)
  
    precedence:
    
      nil > = = true = u > v > del# = . = f# = false = and
    
    partial status:
  
      pi(del#) = []
      pi(.) = []
      pi(f#) = []
      pi(=) = []
      pi(false) = []
      pi(true) = []
      pi(nil) = []
      pi(u) = []
      pi(v) = []
      pi(and) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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#(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          del#_A(x1) = ((0,1),(0,0)) x1 + (1,5)
          ._A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(0,1)) x2 + (15,6)
          f#_A(x1,x2,x3,x4) = ((1,0),(0,0)) x3 + ((0,1),(0,0)) x4 + (8,5)
          =_A(x1,x2) = ((0,1),(0,1)) x1 + (14,15)
          true_A() = (1,4)
          nil_A() = (17,3)
          false_A() = (16,7)
          u_A() = (17,7)
          v_A() = (21,8)
          and_A(x1,x2) = ((0,1),(0,0)) x2 + (1,21)
  
    precedence:
    
      . > f# > del# > = > u > v > true > nil = false = and
    
    partial status:
  
      pi(del#) = []
      pi(.) = []
      pi(f#) = []
      pi(=) = []
      pi(true) = []
      pi(nil) = []
      pi(false) = []
      pi(u) = []
      pi(v) = []
      pi(and) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)