YES

We show the termination of the TRS R:

  a(lambda(x),y) -> lambda(a(x,|1|()))
  a(lambda(x),y) -> lambda(a(x,a(y,t())))
  a(a(x,y),z) -> a(x,a(y,z))
  lambda(x) -> x
  a(x,y) -> x
  a(x,y) -> y

-- SCC decomposition.

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

p1: a#(lambda(x),y) -> lambda#(a(x,|1|()))
p2: a#(lambda(x),y) -> a#(x,|1|())
p3: a#(lambda(x),y) -> lambda#(a(x,a(y,t())))
p4: a#(lambda(x),y) -> a#(x,a(y,t()))
p5: a#(lambda(x),y) -> a#(y,t())
p6: a#(a(x,y),z) -> a#(x,a(y,z))
p7: a#(a(x,y),z) -> a#(y,z)

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> y

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5, p6, p7}


-- Reduction pair.

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

p1: a#(lambda(x),y) -> a#(x,|1|())
p2: a#(a(x,y),z) -> a#(y,z)
p3: a#(a(x,y),z) -> a#(x,a(y,z))
p4: a#(lambda(x),y) -> a#(y,t())
p5: a#(lambda(x),y) -> a#(x,a(y,t()))

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> y

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,3)
          lambda_A(x1) = ((1,0),(0,0)) x1 + (3,2)
          |1|_A() = (0,1)
          a_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (0,2)
          t_A() = (0,6)
  
    precedence:
    
      lambda = |1| = a > a# = t
    
    partial status:
  
      pi(a#) = [1]
      pi(lambda) = []
      pi(|1|) = []
      pi(a) = []
      pi(t) = []

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: a#(lambda(x),y) -> a#(x,|1|())
p2: a#(a(x,y),z) -> a#(y,z)
p3: a#(a(x,y),z) -> a#(x,a(y,z))
p4: a#(lambda(x),y) -> a#(y,t())

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> 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#(lambda(x),y) -> a#(x,|1|())
p2: a#(lambda(x),y) -> a#(y,t())
p3: a#(a(x,y),z) -> a#(x,a(y,z))
p4: a#(a(x,y),z) -> a#(y,z)

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> y

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,2)
          lambda_A(x1) = ((1,0),(0,0)) x1 + (2,1)
          |1|_A() = (0,2)
          t_A() = (0,0)
          a_A(x1,x2) = ((1,0),(1,1)) x1 + x2
  
    precedence:
    
      lambda = |1| = t = a > a#
    
    partial status:
  
      pi(a#) = []
      pi(lambda) = []
      pi(|1|) = []
      pi(t) = []
      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: a#(lambda(x),y) -> a#(y,t())
p2: a#(a(x,y),z) -> a#(x,a(y,z))
p3: a#(a(x,y),z) -> a#(y,z)

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> 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#(lambda(x),y) -> a#(y,t())
p2: a#(a(x,y),z) -> a#(y,z)
p3: a#(a(x,y),z) -> a#(x,a(y,z))

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> y

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,1)
          lambda_A(x1) = ((1,0),(0,0)) x1 + (2,7)
          t_A() = (0,4)
          a_A(x1,x2) = x1 + x2 + (0,2)
          |1|_A() = (0,6)
  
    precedence:
    
      a# = lambda = a > t = |1|
    
    partial status:
  
      pi(a#) = []
      pi(lambda) = []
      pi(t) = []
      pi(a) = []
      pi(|1|) = []

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: a#(a(x,y),z) -> a#(y,z)
p2: a#(a(x,y),z) -> a#(x,a(y,z))

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> y

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> y

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1,x2) = x1 + (1,1)
          a_A(x1,x2) = x1 + x2 + (0,2)
          lambda_A(x1) = ((1,0),(0,0)) x1 + (1,3)
          |1|_A() = (0,6)
          t_A() = (0,0)
  
    precedence:
    
      a# = a = lambda = |1| > t
    
    partial status:
  
      pi(a#) = []
      pi(a) = []
      pi(lambda) = []
      pi(|1|) = []
      pi(t) = []

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#(a(x,y),z) -> a#(y,z)

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> 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#(a(x,y),z) -> a#(y,z)

and R consists of:

r1: a(lambda(x),y) -> lambda(a(x,|1|()))
r2: a(lambda(x),y) -> lambda(a(x,a(y,t())))
r3: a(a(x,y),z) -> a(x,a(y,z))
r4: lambda(x) -> x
r5: a(x,y) -> x
r6: a(x,y) -> y

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1,x2) = ((1,0),(0,0)) x1 + (1,1)
          a_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (2,2)
  
    precedence:
    
      a# = a
    
    partial status:
  
      pi(a#) = []
      pi(a) = [1]

The next rules are strictly ordered:

  p1

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