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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          msort#_A(x1) = max{10, x1 + 2}
          ._A(x1,x2) = max{x1 + 28, x2 + 21}
          del_A(x1,x2) = max{x1 + 13, x2 - 8}
          min_A(x1,x2) = max{x1 + 13, x2 + 1}
          nil_A = 0
          if_A(x1,x2,x3) = max{12, x1 + 2, x2 - 1, x3 - 22}
          <=_A(x1,x2) = max{x1 + 10, x2 + 26}
          =_A(x1,x2) = max{12, x1 + 10, x2 - 1}
    
      precedence:
      
        nil = if > . > msort# = del = min = <= = =
      
      partial status:
    
        pi(msort#) = [1]
        pi(.) = [1, 2]
        pi(del) = [1]
        pi(min) = [2]
        pi(nil) = []
        pi(if) = []
        pi(<=) = [1, 2]
        pi(=) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          msort#_A(x1) = max{56, x1 - 27}
          ._A(x1,x2) = max{55, x1 + 15, x2 + 8}
          del_A(x1,x2) = x1 + 43
          min_A(x1,x2) = max{14, x2 - 18}
          nil_A = 37
          if_A(x1,x2,x3) = 37
          <=_A(x1,x2) = max{x1 + 53, x2 - 2}
          =_A(x1,x2) = x1 + 44
    
      precedence:
      
        msort# = . = del = min = <= > nil = if = =
      
      partial status:
    
        pi(msort#) = []
        pi(.) = [2]
        pi(del) = []
        pi(min) = []
        pi(nil) = []
        pi(if) = []
        pi(<=) = []
        pi(=) = []
    

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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          min#_A(x1,x2) = max{x1 - 2, x2 + 2}
          ._A(x1,x2) = max{x1 - 3, x2}
    
      precedence:
      
        min# = .
      
      partial status:
    
        pi(min#) = [2]
        pi(.) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          min#_A(x1,x2) = x2 + 2
          ._A(x1,x2) = x2 + 1
    
      precedence:
      
        min# = .
      
      partial status:
    
        pi(min#) = []
        pi(.) = [2]
    

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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          del#_A(x1,x2) = max{0, x2 - 2}
          ._A(x1,x2) = max{3, x1, x2 + 1}
    
      precedence:
      
        del# = .
      
      partial status:
    
        pi(del#) = []
        pi(.) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          del#_A(x1,x2) = 0
          ._A(x1,x2) = x2
    
      precedence:
      
        del# = .
      
      partial status:
    
        pi(del#) = []
        pi(.) = [2]
    

The next rules are strictly ordered:

  p1

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