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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1,x2) = x1 + x2
        lambda_A(x1) = x1 + 1
        |1|_A() = 0
        a_A(x1,x2) = x1 + x2
        t_A() = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        a > lambda > |1| > a# > t
      
      argument filter:
    
        pi(a#) = []
        pi(lambda) = []
        pi(|1|) = []
        pi(a) = [1, 2]
        pi(t) = []
    

The next rules are strictly ordered:

  p1, p4, p5

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1,x2) = x1
        a_A(x1,x2) = x1 + x2
        lambda_A(x1) = x1 + 1
        |1|_A() = 0
        t_A() = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        a# > t > |1| > a > lambda
      
      argument filter:
    
        pi(a#) = 1
        pi(a) = [1, 2]
        pi(lambda) = []
        pi(|1|) = []
        pi(t) = []
    

The next rules are strictly ordered:

  p1, p2

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