YES

We show the termination of the TRS R:

  a(h(),h(),h(),x) -> s(x)
  a(l,x,s(y),h()) -> a(l,x,y,s(h()))
  a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
  a(l,s(x),h(),z) -> a(l,x,z,z)
  a(s(l),h(),h(),z) -> a(l,z,h(),z)
  +(x,h()) -> x
  +(h(),x) -> x
  +(s(x),s(y)) -> s(s(+(x,y)))
  +(+(x,y),z) -> +(x,+(y,z))
  s(h()) -> |1|()
  *(h(),x) -> h()
  *(x,h()) -> h()
  *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

-- SCC decomposition.

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

p1: a#(h(),h(),h(),x) -> s#(x)
p2: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p3: a#(l,x,s(y),h()) -> s#(h())
p4: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))
p5: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p6: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p7: a#(s(l),h(),h(),z) -> a#(l,z,h(),z)
p8: +#(s(x),s(y)) -> s#(s(+(x,y)))
p9: +#(s(x),s(y)) -> s#(+(x,y))
p10: +#(s(x),s(y)) -> +#(x,y)
p11: +#(+(x,y),z) -> +#(x,+(y,z))
p12: +#(+(x,y),z) -> +#(y,z)
p13: *#(s(x),s(y)) -> s#(+(+(*(x,y),x),y))
p14: *#(s(x),s(y)) -> +#(+(*(x,y),x),y)
p15: *#(s(x),s(y)) -> +#(*(x,y),x)
p16: *#(s(x),s(y)) -> *#(x,y)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5, p6, p7}
  {p16}
  {p10, p11, p12}


-- Reduction pair.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(s(l),h(),h(),z) -> a#(l,z,h(),z)
p3: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p4: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p5: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      a > s > h > |1| > a#
    
    argument filter:
  
      pi(a#) = 1
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3, 4]
      pi(|1|) = []

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: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p3: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p4: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))
p3: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p4: a#(l,s(x),h(),z) -> a#(l,x,z,z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      a > h > s > |1| > a#
    
    argument filter:
  
      pi(a#) = 2
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3, 4]
      pi(|1|) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))
p3: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p3: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      h > |1| > a > s > a#
    
    argument filter:
  
      pi(a#) = 3
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3, 4]
      pi(|1|) = []

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The set of usable rules consists of

  r10

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      h > s > |1| > a#
    
    argument filter:
  
      pi(a#) = [3, 4]
      pi(s) = [1]
      pi(h) = []
      pi(|1|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > *#
    
    argument filter:
  
      pi(*#) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The set of usable rules consists of

  r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > h > |1| > +# > +
    
    argument filter:
  
      pi(+#) = 1
      pi(s) = 1
      pi(+) = [1, 2]
      pi(h) = []
      pi(|1|) = []

The next rules are strictly ordered:

  p2, p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: *(h(),x) -> h()
r12: *(x,h()) -> h()
r13: *(s(x),s(y)) -> s(+(+(*(x,y),x),y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > +#
    
    argument filter:
  
      pi(+#) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

We remove them from the problem.  Then no dependency pair remains.