YES

We show the termination of the TRS R:

  msort(nil()) -> nil()
  msort(.(x,y)) -> .(min(x,y),msort(del(min(x,y),.(x,y))))
  min(x,nil()) -> x
  min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
  del(x,nil()) -> nil()
  del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))

-- SCC decomposition.

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

p1: msort#(.(x,y)) -> min#(x,y)
p2: msort#(.(x,y)) -> msort#(del(min(x,y),.(x,y)))
p3: msort#(.(x,y)) -> del#(min(x,y),.(x,y))
p4: min#(x,.(y,z)) -> min#(x,z)
p5: min#(x,.(y,z)) -> min#(y,z)
p6: del#(x,.(y,z)) -> del#(x,z)

and R consists of:

r1: msort(nil()) -> nil()
r2: msort(.(x,y)) -> .(min(x,y),msort(del(min(x,y),.(x,y))))
r3: min(x,nil()) -> x
r4: min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
r5: del(x,nil()) -> nil()
r6: del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p4, p5}
  {p6}


-- Reduction pair.

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

p1: msort#(.(x,y)) -> msort#(del(min(x,y),.(x,y)))

and R consists of:

r1: msort(nil()) -> nil()
r2: msort(.(x,y)) -> .(min(x,y),msort(del(min(x,y),.(x,y))))
r3: min(x,nil()) -> x
r4: min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
r5: del(x,nil()) -> nil()
r6: del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))

The set of usable rules consists of

  r3, r4, r5, r6

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      msort#_A(x1) = ((0,0,0),(1,0,0),(0,1,0)) x1
      ._A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(1,1,0),(0,0,1)) x2 + (1,3,0)
      del_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x2 + (0,2,0)
      min_A(x1,x2) = ((1,0,0),(1,0,0),(1,0,0)) x1 + ((1,0,0),(0,1,0),(1,1,0)) x2 + (1,1,1)
      nil_A() = (0,1,2)
      if_A(x1,x2,x3) = ((0,0,0),(0,0,0),(1,0,0)) x1 + x2 + ((0,0,0),(1,0,0),(1,0,0)) x3 + (0,4,4)
      <=_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + ((0,0,0),(0,0,0),(1,0,0)) x2 + (1,1,1)
      =_A(x1,x2) = ((1,0,0),(0,0,0),(1,1,0)) x1 + ((0,0,0),(0,0,0),(1,0,0)) x2 + (1,1,1)

The next rules are strictly ordered:

  p1

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: min#(x,.(y,z)) -> min#(x,z)
p2: min#(x,.(y,z)) -> min#(y,z)

and R consists of:

r1: msort(nil()) -> nil()
r2: msort(.(x,y)) -> .(min(x,y),msort(del(min(x,y),.(x,y))))
r3: min(x,nil()) -> x
r4: min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
r5: del(x,nil()) -> nil()
r6: del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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

and R consists of:

r1: msort(nil()) -> nil()
r2: msort(.(x,y)) -> .(min(x,y),msort(del(min(x,y),.(x,y))))
r3: min(x,nil()) -> x
r4: min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
r5: del(x,nil()) -> nil()
r6: del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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