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, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        towerIter#_A(x1,x2,x3) = x1 + 1
        s_A(x1) = x1 + 17
        p_A(x1) = max{2, x1 - 15}
        exp_A(x1,x2) = max{21, x1 - 18, x2 - 18}
        times_A(x1,x2) = max{4, x1 - 20, x2 - 7}
        |0|_A = 3
        towerIter_A(x1,x2,x3) = max{x1 - 20, x2 + 23, x3 + 1}
  
    base order 2:
  
      precedence:
      
        p > towerIter > exp > |0| > s > towerIter# > times
      
  
    partial status:
  
      pi(towerIter#) = [1]
      pi(s) = [1]
      pi(p) = []
      pi(exp) = []
      pi(times) = []
      pi(|0|) = []
      pi(towerIter) = []

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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        exp#_A(x1,x2) = max{x1 + 14, x2 + 11}
        s_A(x1) = max{12, x1 + 10}
        times_A(x1,x2) = max{27, x1 - 34, x2 - 29}
        |0|_A = 0
  
    base order 2:
  
      precedence:
      
        times > exp# > s > |0|
      
  
    partial status:
  
      pi(exp#) = [2]
      pi(s) = [1]
      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

  r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{1, x1 - 22}
        s_A(x1) = max{40, x1 + 24}
        p_A(x1) = max{12, x1 - 19}
        |0|_A = 3
  
    base order 2:
  
      precedence:
      
        p > s > times# > |0|
      
  
    partial status:
  
      pi(times#) = []
      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: 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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{1, x1 - 16}
        s_A(x1) = x1 + 26
        p_A(x1) = max{13, x1 - 11}
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        p > s > plus# = |0|
      
  
    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, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = max{6, x1 - 12}
        s_A(x1) = max{19, x1 + 11}
        times_A(x1,x2) = max{10, x1 - 46, x2 - 20}
        |0|_A = 7
        towerIter_A(x1,x2,x3) = max{20, x1 - 18, x2 - 2, x3 + 2}
  
    base order 2:
  
      precedence:
      
        towerIter > times > s > p# = |0|
      
  
    partial status:
  
      pi(p#) = []
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []
      pi(towerIter) = []

The following rules are removed:

  p1

Then no dependency pair remains.