YES

We show the termination of the TRS R:

  merge(nil(),y) -> y
  merge(x,nil()) -> x
  merge(.(x,y),.(u,v)) -> if(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))
  ++(nil(),y) -> y
  ++(.(x,y),z) -> .(x,++(y,z))
  if(true(),x,y) -> x
  if(false(),x,y) -> x

-- SCC decomposition.

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

p1: merge#(.(x,y),.(u,v)) -> if#(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))
p2: merge#(.(x,y),.(u,v)) -> merge#(y,.(u,v))
p3: merge#(.(x,y),.(u,v)) -> merge#(.(x,y),v)
p4: ++#(.(x,y),z) -> ++#(y,z)

and R consists of:

r1: merge(nil(),y) -> y
r2: merge(x,nil()) -> x
r3: merge(.(x,y),.(u,v)) -> if(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))
r4: ++(nil(),y) -> y
r5: ++(.(x,y),z) -> .(x,++(y,z))
r6: if(true(),x,y) -> x
r7: if(false(),x,y) -> x

The estimated dependency graph contains the following SCCs:

  {p2, p3}
  {p4}


-- Reduction pair.

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

p1: merge#(.(x,y),.(u,v)) -> merge#(y,.(u,v))
p2: merge#(.(x,y),.(u,v)) -> merge#(.(x,y),v)

and R consists of:

r1: merge(nil(),y) -> y
r2: merge(x,nil()) -> x
r3: merge(.(x,y),.(u,v)) -> if(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))
r4: ++(nil(),y) -> y
r5: ++(.(x,y),z) -> .(x,++(y,z))
r6: if(true(),x,y) -> x
r7: if(false(),x,y) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      merge#_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1 + ((1,0,0),(1,1,0),(1,0,1)) x2
      ._A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(1,0,0),(1,1,0)) x2 + (1,1,1)

The next rules are strictly ordered:

  p1, p2

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

and R consists of:

r1: merge(nil(),y) -> y
r2: merge(x,nil()) -> x
r3: merge(.(x,y),.(u,v)) -> if(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))
r4: ++(nil(),y) -> y
r5: ++(.(x,y),z) -> .(x,++(y,z))
r6: if(true(),x,y) -> x
r7: if(false(),x,y) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      ++#_A(x1,x2) = ((0,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2
      ._A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)

The next rules are strictly ordered:

  p1

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