YES

We show the termination of the TRS R:

  |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
  |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
  |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

-- SCC decomposition.

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

p1: |:|#(|:|(x,y),z) -> |:|#(x,|:|(y,z))
p2: |:|#(|:|(x,y),z) -> |:|#(y,z)
p3: |:|#(+(x,y),z) -> |:|#(x,z)
p4: |:|#(+(x,y),z) -> |:|#(y,z)
p5: |:|#(z,+(x,f(y))) -> |:|#(g(z,y),+(x,a()))

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

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

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |:|#_A(x1,x2) = x1 + 2
        |:|_A(x1,x2) = max{x1 + 3, x2}
        +_A(x1,x2) = max{3, x1, x2}
        f_A(x1) = x1 + 5
        g_A(x1,x2) = x2 + 2
        a_A = 4
  
    precedence:
    
      |:|# = |:| = + = f = g = a
    
    partial status:
  
      pi(|:|#) = []
      pi(|:|) = []
      pi(+) = []
      pi(f) = [1]
      pi(g) = []
      pi(a) = []

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

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

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

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |:|#_A(x1,x2) = max{0, x1 - 2}
        +_A(x1,x2) = max{3, x1, x2 + 1}
        |:|_A(x1,x2) = x2
  
    precedence:
    
      |:|# = + = |:|
    
    partial status:
  
      pi(|:|#) = []
      pi(+) = [2]
      pi(|:|) = [2]

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

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |:|#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
        |:|_A(x1,x2) = x2
        +_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      |:|# = |:| = +
    
    partial status:
  
      pi(|:|#) = [1]
      pi(|:|) = [2]
      pi(+) = [2]

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

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |:|#_A(x1,x2) = max{x1, x2 + 1}
        +_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      |:|# = +
    
    partial status:
  
      pi(|:|#) = [1]
      pi(+) = [2]

The next rules are strictly ordered:

  p1

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