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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        merge#_A(x1,x2) = ((1,1),(1,0)) x1 + ((1,1),(1,1)) x2
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        merge#_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2
        ._A(x1,x2) = ((1,1),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        merge#_A(x1,x2) = ((1,1),(0,1)) x2
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        ++#_A(x1,x2) = ((1,1),(1,1)) x1
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        ++#_A(x1,x2) = ((1,0),(1,0)) x1
        ._A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        ++#_A(x1,x2) = ((1,1),(1,1)) x1
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1

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