YES

We show the termination of the TRS R:

  a(c(d(x))) -> c(x)
  u(b(d(d(x)))) -> b(x)
  v(a(a(x))) -> u(v(x))
  v(a(c(x))) -> u(b(d(x)))
  v(c(x)) -> b(x)
  w(a(a(x))) -> u(w(x))
  w(a(c(x))) -> u(b(d(x)))
  w(c(x)) -> b(x)

-- SCC decomposition.

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

p1: v#(a(a(x))) -> u#(v(x))
p2: v#(a(a(x))) -> v#(x)
p3: v#(a(c(x))) -> u#(b(d(x)))
p4: w#(a(a(x))) -> u#(w(x))
p5: w#(a(a(x))) -> w#(x)
p6: w#(a(c(x))) -> u#(b(d(x)))

and R consists of:

r1: a(c(d(x))) -> c(x)
r2: u(b(d(d(x)))) -> b(x)
r3: v(a(a(x))) -> u(v(x))
r4: v(a(c(x))) -> u(b(d(x)))
r5: v(c(x)) -> b(x)
r6: w(a(a(x))) -> u(w(x))
r7: w(a(c(x))) -> u(b(d(x)))
r8: w(c(x)) -> b(x)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p5}


-- Reduction pair.

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

p1: v#(a(a(x))) -> v#(x)

and R consists of:

r1: a(c(d(x))) -> c(x)
r2: u(b(d(d(x)))) -> b(x)
r3: v(a(a(x))) -> u(v(x))
r4: v(a(c(x))) -> u(b(d(x)))
r5: v(c(x)) -> b(x)
r6: w(a(a(x))) -> u(w(x))
r7: w(a(c(x))) -> u(b(d(x)))
r8: w(c(x)) -> b(x)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      v#_A(x1) = ((1,1),(0,0)) x1
      a_A(x1) = ((1,1),(0,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8

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

-- Reduction pair.

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

p1: w#(a(a(x))) -> w#(x)

and R consists of:

r1: a(c(d(x))) -> c(x)
r2: u(b(d(d(x)))) -> b(x)
r3: v(a(a(x))) -> u(v(x))
r4: v(a(c(x))) -> u(b(d(x)))
r5: v(c(x)) -> b(x)
r6: w(a(a(x))) -> u(w(x))
r7: w(a(c(x))) -> u(b(d(x)))
r8: w(c(x)) -> b(x)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      w#_A(x1) = ((1,1),(0,0)) x1
      a_A(x1) = ((1,1),(0,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8

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