YES

We show the termination of the TRS R:

  |:|(x,x) -> e()
  |:|(x,e()) -> x
  i(|:|(x,y)) -> |:|(y,x)
  |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
  |:|(e(),x) -> i(x)
  i(i(x)) -> x
  i(e()) -> e()
  |:|(x,|:|(y,i(x))) -> i(y)
  |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
  |:|(i(x),|:|(y,x)) -> i(y)
  |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

-- SCC decomposition.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> |:|#(z,i(y))
p4: |:|#(|:|(x,y),z) -> i#(y)
p5: |:|#(e(),x) -> i#(x)
p6: |:|#(x,|:|(y,i(x))) -> i#(y)
p7: |:|#(x,|:|(y,|:|(i(x),z))) -> |:|#(i(z),y)
p8: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p9: |:|#(i(x),|:|(y,x)) -> i#(y)
p10: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p11: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)
p3: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p4: |:|#(i(x),|:|(y,x)) -> i#(y)
p5: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p6: |:|#(x,|:|(y,|:|(i(x),z))) -> |:|#(i(z),y)
p7: |:|#(x,|:|(y,i(x))) -> i#(y)
p8: |:|#(e(),x) -> i#(x)
p9: |:|#(|:|(x,y),z) -> i#(y)
p10: |:|#(|:|(x,y),z) -> |:|#(z,i(y))
p11: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(0,0)) x1 + (3,5)
          |:|_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (4,2)
          |:|#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,4)
          i_A(x1) = ((1,0),(1,1)) x1 + (0,3)
          e_A() = (2,1)
  
    precedence:
    
      i# = |:| = |:|# = i = e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = []
      pi(e) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)
p3: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p4: |:|#(i(x),|:|(y,x)) -> i#(y)
p5: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p6: |:|#(x,|:|(y,i(x))) -> i#(y)
p7: |:|#(e(),x) -> i#(x)
p8: |:|#(|:|(x,y),z) -> i#(y)
p9: |:|#(|:|(x,y),z) -> |:|#(z,i(y))
p10: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> |:|#(z,i(y))
p4: |:|#(|:|(x,y),z) -> i#(y)
p5: |:|#(e(),x) -> i#(x)
p6: |:|#(x,|:|(y,i(x))) -> i#(y)
p7: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p8: |:|#(i(x),|:|(y,x)) -> i#(y)
p9: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p10: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(1,1)) x1 + (5,3)
          |:|_A(x1,x2) = x1 + x2 + (2,1)
          |:|#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (6,4)
          i_A(x1) = x1
          e_A() = (0,0)
  
    precedence:
    
      i# = |:| = |:|# = i = e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = []
      pi(e) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> i#(y)
p4: |:|#(e(),x) -> i#(x)
p5: |:|#(x,|:|(y,i(x))) -> i#(y)
p6: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p7: |:|#(i(x),|:|(y,x)) -> i#(y)
p8: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p9: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)
p3: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p4: |:|#(i(x),|:|(y,x)) -> i#(y)
p5: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p6: |:|#(x,|:|(y,i(x))) -> i#(y)
p7: |:|#(e(),x) -> i#(x)
p8: |:|#(|:|(x,y),z) -> i#(y)
p9: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(0,0)) x1 + (2,5)
          |:|_A(x1,x2) = x1 + x2 + (3,2)
          |:|#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,6)
          i_A(x1) = x1
          e_A() = (1,1)
  
    precedence:
    
      i# = |:| = i > |:|# = e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = []
      pi(e) = []

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: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p3: |:|#(i(x),|:|(y,x)) -> i#(y)
p4: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p5: |:|#(x,|:|(y,i(x))) -> i#(y)
p6: |:|#(e(),x) -> i#(x)
p7: |:|#(|:|(x,y),z) -> i#(y)
p8: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> i#(y)
p4: |:|#(e(),x) -> i#(x)
p5: |:|#(x,|:|(y,i(x))) -> i#(y)
p6: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p7: |:|#(i(x),|:|(y,x)) -> i#(y)
p8: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(0,0)) x1 + (3,4)
          |:|_A(x1,x2) = x1 + x2 + (2,3)
          |:|#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,5)
          i_A(x1) = x1
          e_A() = (2,1)
  
    precedence:
    
      i# = |:| = |:|# = i = e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = [1, 2]
      pi(|:|#) = []
      pi(i) = [1]
      pi(e) = []

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> i#(y)
p4: |:|#(e(),x) -> i#(x)
p5: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p6: |:|#(i(x),|:|(y,x)) -> i#(y)
p7: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p3: |:|#(i(x),|:|(y,x)) -> i#(y)
p4: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p5: |:|#(e(),x) -> i#(x)
p6: |:|#(|:|(x,y),z) -> i#(y)
p7: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          |:|_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (3,0)
          |:|#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (2,3)
          i_A(x1) = ((1,0),(1,1)) x1
          e_A() = (1,1)
  
    precedence:
    
      e > i# = |:|# > |:| = i
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = [1]
      pi(e) = []

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: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p3: |:|#(i(x),|:|(y,x)) -> i#(y)
p4: |:|#(e(),x) -> i#(x)
p5: |:|#(|:|(x,y),z) -> i#(y)
p6: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> i#(y)
p4: |:|#(e(),x) -> i#(x)
p5: |:|#(i(x),|:|(y,x)) -> i#(y)
p6: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((0,0),(1,0)) x1 + (2,4)
          |:|_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,3)
          |:|#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (2,7)
          i_A(x1) = x1 + (0,1)
          e_A() = (1,2)
  
    precedence:
    
      i# = |:| = |:|# = i > e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = []
      pi(e) = []

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: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> i#(y)
p4: |:|#(i(x),|:|(y,x)) -> i#(y)
p5: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p3: |:|#(i(x),|:|(y,x)) -> i#(y)
p4: |:|#(|:|(x,y),z) -> i#(y)
p5: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(1,1)) x1 + (1,0)
          |:|_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,0)
          |:|#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,1)
          i_A(x1) = ((1,0),(1,1)) x1
          e_A() = (1,1)
  
    precedence:
    
      |:| = |:|# = i > i# > e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = []
      pi(e) = []

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: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p3: |:|#(i(x),|:|(y,x)) -> i#(y)
p4: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),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: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(i(x),|:|(y,x)) -> i#(y)
p4: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(1,0)) x1 + (4,0)
          |:|_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,0)
          |:|#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (2,2)
          i_A(x1) = x1 + (0,1)
          e_A() = (0,0)
  
    precedence:
    
      i# = |:| = |:|# = i = e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = []
      pi(e) = []

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

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),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: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,x)) -> i#(y)
p3: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          i#_A(x1) = ((1,0),(0,0)) x1
          |:|_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,1)
          |:|#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (1,2)
          i_A(x1) = x1 + (0,5)
          e_A() = (1,2)
  
    precedence:
    
      i# = |:| = |:|# = i > e
    
    partial status:
  
      pi(i#) = []
      pi(|:|) = []
      pi(|:|#) = []
      pi(i) = []
      pi(e) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: |:|#(i(x),|:|(y,x)) -> i#(y)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          |:|#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,3)
          |:|_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,2)
          i_A(x1) = x1 + (0,1)
          e_A() = (1,3)
  
    precedence:
    
      |:| = i > |:|# = e
    
    partial status:
  
      pi(|:|#) = [1]
      pi(|:|) = []
      pi(i) = []
      pi(e) = []

The next rules are strictly ordered:

  p1

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