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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        msort#_A(x1) = ((0,1),(0,0)) x1
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (0,3)
        del_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(0,0)) x2 + (1,1)
        min_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (1,1)
        nil_A() = (1,1)
        if_A(x1,x2,x3) = ((0,0),(1,0)) x1 + x2
        <=_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        =_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        msort#_A(x1) = (0,0)
        ._A(x1,x2) = x1 + (1,1)
        del_A(x1,x2) = ((1,1),(0,0)) x2 + (1,1)
        min_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
        nil_A() = (1,2)
        if_A(x1,x2,x3) = ((1,1),(0,1)) x2 + (4,2)
        <=_A(x1,x2) = (1,1)
        =_A(x1,x2) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        msort#_A(x1) = (0,0)
        ._A(x1,x2) = ((0,0),(1,0)) x1 + (1,5)
        del_A(x1,x2) = (2,1)
        min_A(x1,x2) = ((1,1),(1,0)) x1 + ((0,1),(0,1)) x2 + (1,1)
        nil_A() = (1,1)
        if_A(x1,x2,x3) = ((1,1),(1,0)) x2 + (3,6)
        <=_A(x1,x2) = (1,1)
        =_A(x1,x2) = (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:

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

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