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^1
        order: lexicographic order
        interpretations:
          del#_A(x1) = x1 + 1
          ._A(x1,x2) = x2 + 6
          f#_A(x1,x2,x3,x4) = x1 + x4 + 7
          =_A(x1,x2) = 4
          false_A() = 1
          true_A() = 3
          nil_A() = 2
          u_A() = 3
          v_A() = 7
          and_A(x1,x2) = x2
  
    precedence:
    
      del# > . = f# = = = false = true = nil = v = and > u
    
    partial status:
  
      pi(del#) = [1]
      pi(.) = [2]
      pi(f#) = []
      pi(=) = []
      pi(false) = []
      pi(true) = []
      pi(nil) = []
      pi(u) = []
      pi(v) = []
      pi(and) = []

The next rules are strictly ordered:

  p3

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#(false(),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#(false(),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^1
        order: lexicographic order
        interpretations:
          del#_A(x1) = x1
          ._A(x1,x2) = x2 + 4
          f#_A(x1,x2,x3,x4) = x4 + 5
          =_A(x1,x2) = x1 + 3
          false_A() = 3
          nil_A() = 1
          true_A() = 0
          u_A() = 2
          v_A() = 5
          and_A(x1,x2) = x2 + 3
  
    precedence:
    
      = = and > del# = . = f# = false = nil = true = u = v
    
    partial status:
  
      pi(del#) = [1]
      pi(.) = [2]
      pi(f#) = [4]
      pi(=) = []
      pi(false) = []
      pi(nil) = []
      pi(true) = []
      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#(false(),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)