YES
We show the termination of the TRS R:

  |0|(|#|()) -> |#|()
  +(|#|(),x) -> x
  +(x,|#|()) -> x
  +(|0|(x),|0|(y)) -> |0|(+(x,y))
  +(|0|(x),|1|(y)) -> |1|(+(x,y))
  +(|1|(x),|0|(y)) -> |1|(+(x,y))
  +(|0|(x),j(y)) -> j(+(x,y))
  +(j(x),|0|(y)) -> j(+(x,y))
  +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
  +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
  +(|1|(x),j(y)) -> |0|(+(x,y))
  +(j(x),|1|(y)) -> |0|(+(x,y))
  +(+(x,y),z) -> +(x,+(y,z))
  opp(|#|()) -> |#|()
  opp(|0|(x)) -> |0|(opp(x))
  opp(|1|(x)) -> j(opp(x))
  opp(j(x)) -> |1|(opp(x))
  -(x,y) -> +(x,opp(y))
  *(|#|(),x) -> |#|()
  *(|0|(x),y) -> |0|(*(x,y))
  *(|1|(x),y) -> +(|0|(*(x,y)),y)
  *(j(x),y) -> -(|0|(*(x,y)),y)
  *(*(x,y),z) -> *(x,*(y,z))


-- SCC decomposition.

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

p1: +#(|0|(x),|0|(y)) -> |0|#(+(x,y))
p2: +#(|0|(x),|0|(y)) -> +#(x,y)
p3: +#(|0|(x),|1|(y)) -> +#(x,y)
p4: +#(|1|(x),|0|(y)) -> +#(x,y)
p5: +#(|0|(x),j(y)) -> +#(x,y)
p6: +#(j(x),|0|(y)) -> +#(x,y)
p7: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p8: +#(|1|(x),|1|(y)) -> +#(x,y)
p9: +#(j(x),j(y)) -> +#(+(x,y),j(|#|()))
p10: +#(j(x),j(y)) -> +#(x,y)
p11: +#(|1|(x),j(y)) -> |0|#(+(x,y))
p12: +#(|1|(x),j(y)) -> +#(x,y)
p13: +#(j(x),|1|(y)) -> |0|#(+(x,y))
p14: +#(j(x),|1|(y)) -> +#(x,y)
p15: +#(+(x,y),z) -> +#(x,+(y,z))
p16: +#(+(x,y),z) -> +#(y,z)
p17: opp#(|0|(x)) -> |0|#(opp(x))
p18: opp#(|0|(x)) -> opp#(x)
p19: opp#(|1|(x)) -> opp#(x)
p20: opp#(j(x)) -> opp#(x)
p21: -#(x,y) -> +#(x,opp(y))
p22: -#(x,y) -> opp#(y)
p23: *#(|0|(x),y) -> |0|#(*(x,y))
p24: *#(|0|(x),y) -> *#(x,y)
p25: *#(|1|(x),y) -> +#(|0|(*(x,y)),y)
p26: *#(|1|(x),y) -> |0|#(*(x,y))
p27: *#(|1|(x),y) -> *#(x,y)
p28: *#(j(x),y) -> -#(|0|(*(x,y)),y)
p29: *#(j(x),y) -> |0|#(*(x,y))
p30: *#(j(x),y) -> *#(x,y)
p31: *#(*(x,y),z) -> *#(x,*(y,z))
p32: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

The estimated dependency graph contains the following SCCs:

  {p24, p27, p30, p31, p32}
  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p12, p14, p15, p16}
  {p18, p19, p20}


-- Reduction pair.

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

p1: *#(*(x,y),z) -> *#(y,z)
p2: *#(*(x,y),z) -> *#(x,*(y,z))
p3: *#(j(x),y) -> *#(x,y)
p4: *#(|1|(x),y) -> *#(x,y)
p5: *#(|0|(x),y) -> *#(x,y)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r2, r3, r14, r19

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:
              *#_A(x1,x2) = x1
              *_A(x1,x2) = x1 + x2 + 1
              j_A(x1) = x1 + 3
              |1|_A(x1) = x1 + 4
              |0|_A(x1) = x1 + 1
              +_A(x1,x2) = x1 + x2 + 4
              |#|_A() = 1
              opp_A(x1) = x1 + 1
              *#!_A(x1,x2) = x1
              *!_A(x1,x2) = x1 + 2
              j!_A(x1) = x1 + 10
              |1|!_A(x1) = 9
              |0|!_A(x1) = x1 + 13
              +!_A(x1,x2) = x2 + 7
              |#|!_A() = 0
              opp!_A(x1) = 12
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = 0
            *_A(x1,x2) = max{x1 + 59, x2 + 59}
            j_A(x1) = max{48, x1 + 21}
            |1|_A(x1) = max{57, x1 - 55}
            |0|_A(x1) = max{0, x1 - 12}
            +_A(x1,x2) = max{19, x1 + 5, x2 + 6}
            |#|_A = 22
            opp_A(x1) = max{56, x1 - 58}
            *#!_A(x1,x2) = 3
            *!_A(x1,x2) = max{0, x1 - 58}
            j!_A(x1) = x1 + 5
            |1|!_A(x1) = 4
            |0|!_A(x1) = 38
            +!_A(x1,x2) = max{0, x2 - 20}
            |#|!_A = 0
            opp!_A(x1) = 0
        
  
    partial status:
  
      pi(*#) = []
      pi(*) = []
      pi(j) = [1]
      pi(|1|) = []
      pi(|0|) = [1]
      pi(+) = [1, 2]
      pi(|#|) = []
      pi(opp) = [1]

The following rules are removed:

  p1, p2, p3, p4, p5

Then no dependency pair remains.

-- Reduction pair.

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

p1: +#(|0|(x),|0|(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))
p4: +#(j(x),|1|(y)) -> +#(x,y)
p5: +#(|1|(x),j(y)) -> +#(x,y)
p6: +#(j(x),j(y)) -> +#(x,y)
p7: +#(j(x),j(y)) -> +#(+(x,y),j(|#|()))
p8: +#(|0|(x),j(y)) -> +#(x,y)
p9: +#(|1|(x),|1|(y)) -> +#(x,y)
p10: +#(|1|(x),|1|(y)) -> +#(+(x,y),|1|(|#|()))
p11: +#(|0|(x),|1|(y)) -> +#(x,y)
p12: +#(j(x),|0|(y)) -> +#(x,y)
p13: +#(|1|(x),|0|(y)) -> +#(x,y)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r19

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:
              +#_A(x1,x2) = 0
              |0|_A(x1) = x1
              +_A(x1,x2) = x1 + x2
              j_A(x1) = x1 + 2
              |1|_A(x1) = x1 + 2
              |#|_A() = 0
              opp_A(x1) = 1
              *_A(x1,x2) = x1 + x2 + 1
              +#!_A(x1,x2) = x1 + x2
              |0|!_A(x1) = 3
              +!_A(x1,x2) = 3
              j!_A(x1) = 3
              |1|!_A(x1) = 3
              |#|!_A() = 1
              opp!_A(x1) = 2
              *!_A(x1,x2) = x1 + 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = 0
            |0|_A(x1) = max{8, x1}
            +_A(x1,x2) = max{6, x1, x2}
            j_A(x1) = max{11, x1}
            |1|_A(x1) = max{11, x1}
            |#|_A = 11
            opp_A(x1) = 14
            *_A(x1,x2) = max{10, x1 + 1}
            +#!_A(x1,x2) = max{19, x1 + 12, x2 + 12}
            |0|!_A(x1) = 0
            +!_A(x1,x2) = 0
            j!_A(x1) = 0
            |1|!_A(x1) = 0
            |#|!_A = 0
            opp!_A(x1) = max{1, x1 - 12}
            *!_A(x1,x2) = max{21, x1 + 11, x2 + 11}
        
  
    partial status:
  
      pi(+#) = []
      pi(|0|) = []
      pi(+) = [1, 2]
      pi(j) = []
      pi(|1|) = []
      pi(|#|) = []
      pi(opp) = []
      pi(*) = []

The following rules are removed:

  p4, p5, p6, p7, p8, p9, p10, p11, p12, p13



-- SCC decomposition.

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

p1: +#(|0|(x),|0|(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

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: +#(|0|(x),|0|(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(x,+(y,z))
p3: +#(+(x,y),z) -> +#(y,z)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r19

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:
              +#_A(x1,x2) = x1 + x2
              |0|_A(x1) = x1 + 8
              +_A(x1,x2) = x1 + x2
              |#|_A() = 0
              |1|_A(x1) = x1 + 7
              j_A(x1) = x1 + 7
              opp_A(x1) = x1 + 1
              *_A(x1,x2) = x1 + x2 + 4
              +#!_A(x1,x2) = x1
              |0|!_A(x1) = x1 + 12
              +!_A(x1,x2) = x1 + x2 + 1
              |#|!_A() = 0
              |1|!_A(x1) = 14
              j!_A(x1) = 13
              opp!_A(x1) = 11
              *!_A(x1,x2) = 15
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +#_A(x1,x2) = 0
            |0|_A(x1) = max{12, x1 - 6}
            +_A(x1,x2) = max{19, x1, x2}
            |#|_A = 0
            |1|_A(x1) = max{11, x1 - 6}
            j_A(x1) = max{10, x1 - 6}
            opp_A(x1) = max{9, x1 - 20}
            *_A(x1,x2) = max{7, x1 - 23, x2 - 23}
            +#!_A(x1,x2) = max{1, x1, x2 - 20}
            |0|!_A(x1) = max{8, x1 + 2}
            +!_A(x1,x2) = 3
            |#|!_A = 3
            |1|!_A(x1) = x1 + 3
            j!_A(x1) = x1 + 5
            opp!_A(x1) = max{3, x1 - 13}
            *!_A(x1,x2) = 0
        
  
    partial status:
  
      pi(+#) = [1]
      pi(|0|) = [1]
      pi(+) = [1, 2]
      pi(|#|) = []
      pi(|1|) = [1]
      pi(j) = [1]
      pi(opp) = [1]
      pi(*) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: opp#(|0|(x)) -> opp#(x)
p2: opp#(j(x)) -> opp#(x)
p3: opp#(|1|(x)) -> opp#(x)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r1, r19

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:
              opp#_A(x1) = x1
              |0|_A(x1) = x1
              j_A(x1) = x1 + 1
              |1|_A(x1) = x1 + 1
              |#|_A() = 2
              *_A(x1,x2) = x1 + x2 + 1
              opp#!_A(x1) = x1 + 8
              |0|!_A(x1) = 1
              j!_A(x1) = x1 + 7
              |1|!_A(x1) = 2
              |#|!_A() = 0
              *!_A(x1,x2) = x1 + 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            opp#_A(x1) = 0
            |0|_A(x1) = max{51, x1}
            j_A(x1) = max{1, x1}
            |1|_A(x1) = max{27, x1 + 9}
            |#|_A = 0
            *_A(x1,x2) = max{1, x1 - 82, x2 - 95}
            opp#!_A(x1) = max{4, x1 - 52}
            |0|!_A(x1) = max{56, x1 + 3}
            j!_A(x1) = 1
            |1|!_A(x1) = 6
            |#|!_A = 2
            *!_A(x1,x2) = max{43, x1 - 94, x2 - 98}
        
  
    partial status:
  
      pi(opp#) = []
      pi(|0|) = [1]
      pi(j) = []
      pi(|1|) = [1]
      pi(|#|) = []
      pi(*) = [1, 2]

The following rules are removed:

  p2, p3



-- SCC decomposition.

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

p1: opp#(|0|(x)) -> opp#(x)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: opp#(|0|(x)) -> opp#(x)

and R consists of:

r1: |0|(|#|()) -> |#|()
r2: +(|#|(),x) -> x
r3: +(x,|#|()) -> x
r4: +(|0|(x),|0|(y)) -> |0|(+(x,y))
r5: +(|0|(x),|1|(y)) -> |1|(+(x,y))
r6: +(|1|(x),|0|(y)) -> |1|(+(x,y))
r7: +(|0|(x),j(y)) -> j(+(x,y))
r8: +(j(x),|0|(y)) -> j(+(x,y))
r9: +(|1|(x),|1|(y)) -> j(+(+(x,y),|1|(|#|())))
r10: +(j(x),j(y)) -> |1|(+(+(x,y),j(|#|())))
r11: +(|1|(x),j(y)) -> |0|(+(x,y))
r12: +(j(x),|1|(y)) -> |0|(+(x,y))
r13: +(+(x,y),z) -> +(x,+(y,z))
r14: opp(|#|()) -> |#|()
r15: opp(|0|(x)) -> |0|(opp(x))
r16: opp(|1|(x)) -> j(opp(x))
r17: opp(j(x)) -> |1|(opp(x))
r18: -(x,y) -> +(x,opp(y))
r19: *(|#|(),x) -> |#|()
r20: *(|0|(x),y) -> |0|(*(x,y))
r21: *(|1|(x),y) -> +(|0|(*(x,y)),y)
r22: *(j(x),y) -> -(|0|(*(x,y)),y)
r23: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r1, r19

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:
              opp#_A(x1) = x1
              |0|_A(x1) = x1 + 9
              |#|_A() = 7
              *_A(x1,x2) = x1 + x2 + 1
              opp#!_A(x1) = x1
              |0|!_A(x1) = 4
              |#|!_A() = 0
              *!_A(x1,x2) = x1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            opp#_A(x1) = 0
            |0|_A(x1) = max{23, x1 - 25}
            |#|_A = 8
            *_A(x1,x2) = max{14, x1 - 56, x2 - 49}
            opp#!_A(x1) = max{5, x1 - 24}
            |0|!_A(x1) = max{4, x1 - 9}
            |#|!_A = 1
            *!_A(x1,x2) = max{6, x1 - 39, x2 - 52}
        
  
    partial status:
  
      pi(opp#) = [1]
      pi(|0|) = [1]
      pi(|#|) = []
      pi(*) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.