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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      a#_A(x1,x2) = ((1,0,0),(0,1,0),(0,1,1)) x1 + x2
      lambda_A(x1) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (1,2,1)
      |1|_A() = (0,1,1)
      a_A(x1,x2) = x1 + x2 + (0,2,1)
      t_A() = (0,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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