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

  r1, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              towerIter#_A(x1,x2,x3) = 0
              s_A(x1) = 2
              p_A(x1) = 6
              exp_A(x1,x2) = x1 + 1
              plus_A(x1,x2) = x2 + 1
              |0|_A() = 6
              towerIter#!_A(x1,x2,x3) = 2
              s!_A(x1) = 6
              p!_A(x1) = 6
              exp!_A(x1,x2) = x1 + x2 + 1
              plus!_A(x1,x2) = 0
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            towerIter#_A(x1,x2,x3) = max{0, x1 - 47, x2 - 28}
            s_A(x1) = x1 + 46
            p_A(x1) = max{32, x1 - 14}
            exp_A(x1,x2) = max{33, x1 + 5, x2 + 1}
            plus_A(x1,x2) = max{105, x2 + 15}
            |0|_A = 104
            towerIter#!_A(x1,x2,x3) = x2 + 2
            s!_A(x1) = max{67, x1 + 34}
            p!_A(x1) = max{67, x1 - 24}
            exp!_A(x1,x2) = 0
            plus!_A(x1,x2) = x2 + 50
            |0|!_A = 68
        
  
    partial status:
  
      pi(towerIter#) = [1]
      pi(s) = [1]
      pi(p) = []
      pi(exp) = []
      pi(plus) = [2]
      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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              exp#_A(x1,x2) = x2
              s_A(x1) = x1 + 3
              exp#!_A(x1,x2) = 0
              s!_A(x1) = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            exp#_A(x1,x2) = 0
            s_A(x1) = max{2, x1 - 7}
            exp#!_A(x1,x2) = max{0, x2 - 4}
            s!_A(x1) = max{22, x1 - 3}
        
  
    partial status:
  
      pi(exp#) = [2]
      pi(s) = [1]

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, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              times#_A(x1,x2) = x1
              s_A(x1) = x1 + 3
              p_A(x1) = x1
              |0|_A() = 6
              times#!_A(x1,x2) = 0
              s!_A(x1) = 5
              p!_A(x1) = x1
              |0|!_A() = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 22, x2 - 24}
            s_A(x1) = x1 + 21
            p_A(x1) = max{4, x1 - 17}
            |0|_A = 21
            times#!_A(x1,x2) = max{8, x1 + 3}
            s!_A(x1) = x1 + 9
            p!_A(x1) = 60
            |0|!_A = 31
        
  
    partial status:
  
      pi(times#) = []
      pi(s) = []
      pi(p) = [1]
      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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = x1 + 2
              p_A(x1) = x1
              |0|_A() = 5
              plus#!_A(x1,x2) = x1 + 8
              s!_A(x1) = x1 + 2
              p!_A(x1) = 10
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 21, x2 - 2}
            s_A(x1) = max{35, x1 + 22}
            p_A(x1) = max{20, x1 - 10}
            |0|_A = 64
            plus#!_A(x1,x2) = max{25, x1 + 6}
            s!_A(x1) = max{24, x1 - 36}
            p!_A(x1) = max{23, x1 - 11}
            |0|!_A = 26
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(p) = [1]
      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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              p#_A(x1) = x1
              s_A(x1) = x1 + 3
              p#!_A(x1) = x1 + 2
              s!_A(x1) = x1 + 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            p#_A(x1) = 0
            s_A(x1) = max{22, x1 - 21}
            p#!_A(x1) = 0
            s!_A(x1) = max{10, x1 + 1}
        
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.