YES
We show the termination of the TRS R:

  plus(|0|(),x) -> x
  plus(s(x),y) -> s(plus(p(s(x)),y))
  times(|0|(),y) -> |0|()
  times(s(x),y) -> plus(y,times(p(s(x)),y))
  exp(x,|0|()) -> s(|0|())
  exp(x,s(y)) -> times(x,exp(x,y))
  p(s(|0|())) -> |0|()
  p(s(s(x))) -> s(p(s(x)))
  tower(x,y) -> towerIter(x,y,s(|0|()))
  towerIter(|0|(),y,z) -> z
  towerIter(s(x),y,z) -> towerIter(p(s(x)),y,exp(y,z))


-- SCC decomposition.

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

p1: plus#(s(x),y) -> plus#(p(s(x)),y)
p2: plus#(s(x),y) -> p#(s(x))
p3: times#(s(x),y) -> plus#(y,times(p(s(x)),y))
p4: times#(s(x),y) -> times#(p(s(x)),y)
p5: times#(s(x),y) -> p#(s(x))
p6: exp#(x,s(y)) -> times#(x,exp(x,y))
p7: exp#(x,s(y)) -> exp#(x,y)
p8: p#(s(s(x))) -> p#(s(x))
p9: tower#(x,y) -> towerIter#(x,y,s(|0|()))
p10: towerIter#(s(x),y,z) -> towerIter#(p(s(x)),y,exp(y,z))
p11: towerIter#(s(x),y,z) -> p#(s(x))
p12: towerIter#(s(x),y,z) -> exp#(y,z)

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: exp(x,|0|()) -> s(|0|())
r6: exp(x,s(y)) -> times(x,exp(x,y))
r7: p(s(|0|())) -> |0|()
r8: p(s(s(x))) -> s(p(s(x)))
r9: tower(x,y) -> towerIter(x,y,s(|0|()))
r10: towerIter(|0|(),y,z) -> z
r11: towerIter(s(x),y,z) -> towerIter(p(s(x)),y,exp(y,z))

The estimated dependency graph contains the following SCCs:

  {p10}
  {p7}
  {p4}
  {p1}
  {p8}


-- Reduction pair.

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

p1: towerIter#(s(x),y,z) -> towerIter#(p(s(x)),y,exp(y,z))

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: exp(x,|0|()) -> s(|0|())
r6: exp(x,s(y)) -> times(x,exp(x,y))
r7: p(s(|0|())) -> |0|()
r8: p(s(s(x))) -> s(p(s(x)))
r9: tower(x,y) -> towerIter(x,y,s(|0|()))
r10: towerIter(|0|(),y,z) -> z
r11: towerIter(s(x),y,z) -> towerIter(p(s(x)),y,exp(y,z))

The set of usable rules consists of

  r3, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        towerIter#_A(x1,x2,x3) = x2 + 20
        s_A(x1) = max{18, x1 + 8}
        p_A(x1) = max{14, x1}
        exp_A(x1,x2) = 1
        times_A(x1,x2) = max{1, x1 - 10, x2 - 2}
        |0|_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          towerIter#_A(x1,x2,x3) = max{0, x1 - 28, x2 - 2}
          s_A(x1) = max{48, x1 + 23}
          p_A(x1) = max{8, x1 - 21}
          exp_A(x1,x2) = 3
          times_A(x1,x2) = max{4, x1 + 2, x2 - 1}
          |0|_A = 46
          towerIter#!_A(x1,x2,x3) = max{12, x1 - 10}
          s!_A(x1) = max{18, x1 + 11}
          p!_A(x1) = 10
          exp!_A(x1,x2) = max{58, x1 - 8, x2 - 47}
          times!_A(x1,x2) = max{8, x1 - 8}
          |0|!_A = 9
  
    partial status:
  
      pi(towerIter#) = [2]
      pi(s) = [1]
      pi(p) = [1]
      pi(exp) = []
      pi(times) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: exp#(x,s(y)) -> exp#(x,y)

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: exp(x,|0|()) -> s(|0|())
r6: exp(x,s(y)) -> times(x,exp(x,y))
r7: p(s(|0|())) -> |0|()
r8: p(s(s(x))) -> s(p(s(x)))
r9: tower(x,y) -> towerIter(x,y,s(|0|()))
r10: towerIter(|0|(),y,z) -> z
r11: towerIter(s(x),y,z) -> towerIter(p(s(x)),y,exp(y,z))

The set of usable rules consists of

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        exp#_A(x1,x2) = max{18, x1 - 1, x2 + 10}
        s_A(x1) = max{17, x1 + 9}
        times_A(x1,x2) = max{28, x1 - 28, x2 - 31}
        |0|_A = 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          exp#_A(x1,x2) = 0
          s_A(x1) = max{18, x1 - 19}
          times_A(x1,x2) = max{24, x1 - 20, x2 - 49}
          |0|_A = 18
          exp#!_A(x1,x2) = 0
          s!_A(x1) = max{4, x1 - 95}
          times!_A(x1,x2) = max{2, x1 - 40, x2 - 49}
          |0|!_A = 0
  
    partial status:
  
      pi(exp#) = [2]
      pi(s) = []
      pi(times) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: times#(s(x),y) -> times#(p(s(x)),y)

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: exp(x,|0|()) -> s(|0|())
r6: exp(x,s(y)) -> times(x,exp(x,y))
r7: p(s(|0|())) -> |0|()
r8: p(s(s(x))) -> s(p(s(x)))
r9: tower(x,y) -> towerIter(x,y,s(|0|()))
r10: towerIter(|0|(),y,z) -> z
r11: towerIter(s(x),y,z) -> towerIter(p(s(x)),y,exp(y,z))

The set of usable rules consists of

  r3, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{x1 + 3, x2 + 11}
        s_A(x1) = 6
        p_A(x1) = 8
        times_A(x1,x2) = 0
        |0|_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = max{0, x1 - 20, x2 - 20}
          s_A(x1) = x1 + 19
          p_A(x1) = max{15, x1 - 4}
          times_A(x1,x2) = max{8, x1 - 113, x2 - 122}
          |0|_A = 2
          times#!_A(x1,x2) = max{2, x1 - 16}
          s!_A(x1) = max{1, x1 - 54}
          p!_A(x1) = max{5, x1 - 39}
          times!_A(x1,x2) = max{3, x1 - 125, x2 - 109}
          |0|!_A = 0
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = []
      pi(p) = []
      pi(times) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: plus#(s(x),y) -> plus#(p(s(x)),y)

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: exp(x,|0|()) -> s(|0|())
r6: exp(x,s(y)) -> times(x,exp(x,y))
r7: p(s(|0|())) -> |0|()
r8: p(s(s(x))) -> s(p(s(x)))
r9: tower(x,y) -> towerIter(x,y,s(|0|()))
r10: towerIter(|0|(),y,z) -> z
r11: towerIter(s(x),y,z) -> towerIter(p(s(x)),y,exp(y,z))

The set of usable rules consists of

  r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{0, x1 - 16}
        s_A(x1) = max{32, x1 + 22}
        p_A(x1) = max{17, x1 - 13}
        |0|_A = 20
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 39, x2 - 57}
          s_A(x1) = max{0, x1 - 39}
          p_A(x1) = max{38, x1 - 17}
          |0|_A = 27
          plus#!_A(x1,x2) = x2 + 13
          s!_A(x1) = max{14, x1 - 75}
          p!_A(x1) = x1 + 98
          |0|!_A = 95
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(p) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: exp(x,|0|()) -> s(|0|())
r6: exp(x,s(y)) -> times(x,exp(x,y))
r7: p(s(|0|())) -> |0|()
r8: p(s(s(x))) -> s(p(s(x)))
r9: tower(x,y) -> towerIter(x,y,s(|0|()))
r10: towerIter(|0|(),y,z) -> z
r11: towerIter(s(x),y,z) -> towerIter(p(s(x)),y,exp(y,z))

The set of usable rules consists of

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = max{32, x1}
        s_A(x1) = max{21, x1 + 11}
        times_A(x1,x2) = max{1, x1 - 29, x2 - 49}
        |0|_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          p#_A(x1) = 0
          s_A(x1) = max{5, x1 - 4}
          times_A(x1,x2) = max{31, x1 - 8, x2 - 3}
          |0|_A = 3
          p#!_A(x1) = x1 + 10
          s!_A(x1) = max{16, x1 + 11}
          times!_A(x1,x2) = max{19, x1 - 52, x2 - 41}
          |0|!_A = 18
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.