YES

We show the termination of the TRS R:

  eq(|0|(),|0|()) -> true()
  eq(|0|(),s(x)) -> false()
  eq(s(x),|0|()) -> false()
  eq(s(x),s(y)) -> eq(x,y)
  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  app(nil(),y) -> y
  app(add(n,x),y) -> add(n,app(x,y))
  min(add(n,nil())) -> n
  min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
  if_min(true(),add(n,add(m,x))) -> min(add(n,x))
  if_min(false(),add(n,add(m,x))) -> min(add(m,x))
  rm(n,nil()) -> nil()
  rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
  if_rm(true(),n,add(m,x)) -> rm(n,x)
  if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
  minsort(nil(),nil()) -> nil()
  minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
  if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
  if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

-- SCC decomposition.

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

p1: eq#(s(x),s(y)) -> eq#(x,y)
p2: le#(s(x),s(y)) -> le#(x,y)
p3: app#(add(n,x),y) -> app#(x,y)
p4: min#(add(n,add(m,x))) -> if_min#(le(n,m),add(n,add(m,x)))
p5: min#(add(n,add(m,x))) -> le#(n,m)
p6: if_min#(true(),add(n,add(m,x))) -> min#(add(n,x))
p7: if_min#(false(),add(n,add(m,x))) -> min#(add(m,x))
p8: rm#(n,add(m,x)) -> if_rm#(eq(n,m),n,add(m,x))
p9: rm#(n,add(m,x)) -> eq#(n,m)
p10: if_rm#(true(),n,add(m,x)) -> rm#(n,x)
p11: if_rm#(false(),n,add(m,x)) -> rm#(n,x)
p12: minsort#(add(n,x),y) -> if_minsort#(eq(n,min(add(n,x))),add(n,x),y)
p13: minsort#(add(n,x),y) -> eq#(n,min(add(n,x)))
p14: minsort#(add(n,x),y) -> min#(add(n,x))
p15: if_minsort#(true(),add(n,x),y) -> minsort#(app(rm(n,x),y),nil())
p16: if_minsort#(true(),add(n,x),y) -> app#(rm(n,x),y)
p17: if_minsort#(true(),add(n,x),y) -> rm#(n,x)
p18: if_minsort#(false(),add(n,x),y) -> minsort#(x,add(n,y))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The estimated dependency graph contains the following SCCs:

  {p12, p15, p18}
  {p8, p10, p11}
  {p1}
  {p4, p6, p7}
  {p2}
  {p3}


-- Reduction pair.

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

p1: if_minsort#(false(),add(n,x),y) -> minsort#(x,add(n,y))
p2: minsort#(add(n,x),y) -> if_minsort#(eq(n,min(add(n,x))),add(n,x),y)
p3: if_minsort#(true(),add(n,x),y) -> minsort#(app(rm(n,x),y),nil())

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            if_minsort#_A(x1,x2,x3) = ((0,1),(0,0)) x1 + ((0,1),(0,1)) x2 + ((0,1),(0,1)) x3 + (1,5)
            false_A() = (1,2)
            add_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(0,1)) x2 + (0,4)
            minsort#_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (3,5)
            eq_A(x1,x2) = (1,2)
            min_A(x1) = ((0,1),(1,0)) x1 + (2,9)
            true_A() = (1,1)
            app_A(x1,x2) = x1 + ((1,1),(0,1)) x2
            rm_A(x1,x2) = ((1,1),(1,0)) x1 + ((0,1),(0,1)) x2 + (0,1)
            nil_A() = (0,1)
            le_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (3,9)
            |0|_A() = (2,2)
            s_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            if_min_A(x1,x2) = ((0,1),(1,0)) x2 + (1,5)
            if_rm_A(x1,x2,x3) = ((0,1),(1,0)) x1 + ((1,1),(1,0)) x2 + x3 + (1,0)
    
      precedence:
      
        |0| > rm = if_rm > if_minsort# = minsort# > eq > true > add = app > min > nil > le = if_min > false > s
      
      partial status:
    
        pi(if_minsort#) = []
        pi(false) = []
        pi(add) = []
        pi(minsort#) = []
        pi(eq) = []
        pi(min) = []
        pi(true) = []
        pi(app) = [1, 2]
        pi(rm) = []
        pi(nil) = []
        pi(le) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(if_min) = []
        pi(if_rm) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            if_minsort#_A(x1,x2,x3) = (0,2)
            false_A() = (3,0)
            add_A(x1,x2) = (5,3)
            minsort#_A(x1,x2) = (0,2)
            eq_A(x1,x2) = (6,5)
            min_A(x1) = (2,4)
            true_A() = (0,1)
            app_A(x1,x2) = (6,4)
            rm_A(x1,x2) = (4,4)
            nil_A() = (1,3)
            le_A(x1,x2) = (7,0)
            |0|_A() = (1,0)
            s_A(x1) = ((1,1),(1,1)) x1 + (7,5)
            if_min_A(x1,x2) = (1,4)
            if_rm_A(x1,x2,x3) = (4,4)
    
      precedence:
      
        rm = if_min = if_rm > false > le = s > eq = app > add = min > true > nil > |0| > if_minsort# = minsort#
      
      partial status:
    
        pi(if_minsort#) = []
        pi(false) = []
        pi(add) = []
        pi(minsort#) = []
        pi(eq) = []
        pi(min) = []
        pi(true) = []
        pi(app) = []
        pi(rm) = []
        pi(nil) = []
        pi(le) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(if_min) = []
        pi(if_rm) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: if_minsort#(false(),add(n,x),y) -> minsort#(x,add(n,y))
p2: minsort#(add(n,x),y) -> if_minsort#(eq(n,min(add(n,x))),add(n,x),y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: if_minsort#(false(),add(n,x),y) -> minsort#(x,add(n,y))
p2: minsort#(add(n,x),y) -> if_minsort#(eq(n,min(add(n,x))),add(n,x),y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r10, r11, r12, r13

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minsort#_A(x1,x2,x3) = max{51, x1 - 2, x2 + 38, x3 + 24}
          false_A = 16
          add_A(x1,x2) = max{13, x1 + 12, x2}
          minsort#_A(x1,x2) = max{x1 + 38, x2 + 24}
          eq_A(x1,x2) = max{x1 + 52, x2 + 10}
          min_A(x1) = max{1, x1}
          le_A(x1,x2) = max{x1 + 11, x2 + 2}
          |0|_A = 17
          true_A = 1
          s_A(x1) = max{8, x1 + 7}
          if_min_A(x1,x2) = max{x1 + 1, x2}
          nil_A = 0
    
      precedence:
      
        eq > add > min = le > false = |0| = s = if_min = nil > if_minsort# = minsort# = true
      
      partial status:
    
        pi(if_minsort#) = [2]
        pi(false) = []
        pi(add) = [1, 2]
        pi(minsort#) = [1]
        pi(eq) = []
        pi(min) = [1]
        pi(le) = [1, 2]
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
        pi(if_min) = [1, 2]
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minsort#_A(x1,x2,x3) = max{11, x2 - 6}
          false_A = 32
          add_A(x1,x2) = max{23, x1 + 21, x2 + 17}
          minsort#_A(x1,x2) = max{10, x1 + 1}
          eq_A(x1,x2) = 25
          min_A(x1) = max{6, x1 - 14}
          le_A(x1,x2) = max{4, x1 + 2}
          |0|_A = 25
          true_A = 26
          s_A(x1) = x1 + 33
          if_min_A(x1,x2) = max{21, x2 - 14}
          nil_A = 0
    
      precedence:
      
        nil > eq = true > min = if_min > if_minsort# = |0| = s > false = add = minsort# = le
      
      partial status:
    
        pi(if_minsort#) = []
        pi(false) = []
        pi(add) = [1]
        pi(minsort#) = [1]
        pi(eq) = []
        pi(min) = []
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
        pi(if_min) = []
        pi(nil) = []
    

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: if_rm#(false(),n,add(m,x)) -> rm#(n,x)
p2: rm#(n,add(m,x)) -> if_rm#(eq(n,m),n,add(m,x))
p3: if_rm#(true(),n,add(m,x)) -> rm#(n,x)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_rm#_A(x1,x2,x3) = max{x1 + 4, x2 + 18, x3}
          false_A = 7
          add_A(x1,x2) = max{x1, x2 + 9}
          rm#_A(x1,x2) = max{x1 + 18, x2 + 9}
          eq_A(x1,x2) = max{x1 + 14, x2 + 4}
          true_A = 9
          |0|_A = 6
          s_A(x1) = x1 + 8
    
      precedence:
      
        false = eq = true > add > if_rm# = rm# = |0| = s
      
      partial status:
    
        pi(if_rm#) = [3]
        pi(false) = []
        pi(add) = [1]
        pi(rm#) = [2]
        pi(eq) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_rm#_A(x1,x2,x3) = x3
          false_A = 6
          add_A(x1,x2) = max{4, x1 + 2}
          rm#_A(x1,x2) = max{1, x2}
          eq_A(x1,x2) = 5
          true_A = 6
          |0|_A = 0
          s_A(x1) = x1 + 7
    
      precedence:
      
        false = add = true = s > eq > if_rm# = rm# = |0|
      
      partial status:
    
        pi(if_rm#) = []
        pi(false) = []
        pi(add) = [1]
        pi(rm#) = [2]
        pi(eq) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: rm#(n,add(m,x)) -> if_rm#(eq(n,m),n,add(m,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: eq#(s(x),s(y)) -> eq#(x,y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

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:
        
          eq#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        eq# = s
      
      partial status:
    
        pi(eq#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          eq#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        eq# = s
      
      partial status:
    
        pi(eq#) = []
        pi(s) = [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: if_min#(false(),add(n,add(m,x))) -> min#(add(m,x))
p2: min#(add(n,add(m,x))) -> if_min#(le(n,m),add(n,add(m,x)))
p3: if_min#(true(),add(n,add(m,x))) -> min#(add(n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_min#_A(x1,x2) = max{x1 + 5, x2}
          false_A = 4
          add_A(x1,x2) = max{x1 + 9, x2 + 3}
          min#_A(x1) = x1
          le_A(x1,x2) = x2 + 2
          true_A = 1
          |0|_A = 5
          s_A(x1) = x1 + 8
    
      precedence:
      
        if_min# = false = add = min# = le = true = |0| = s
      
      partial status:
    
        pi(if_min#) = []
        pi(false) = []
        pi(add) = []
        pi(min#) = []
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_min#_A(x1,x2) = 15
          false_A = 18
          add_A(x1,x2) = 19
          min#_A(x1) = max{14, x1 - 4}
          le_A(x1,x2) = 13
          true_A = 22
          |0|_A = 17
          s_A(x1) = x1 + 19
    
      precedence:
      
        if_min# = false = add = min# = le = true = |0| = s
      
      partial status:
    
        pi(if_min#) = []
        pi(false) = []
        pi(add) = []
        pi(min#) = []
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: min#(add(n,add(m,x))) -> if_min#(le(n,m),add(n,add(m,x)))
p2: if_min#(true(),add(n,add(m,x))) -> min#(add(n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: min#(add(n,add(m,x))) -> if_min#(le(n,m),add(n,add(m,x)))
p2: if_min#(true(),add(n,add(m,x))) -> min#(add(n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          min#_A(x1) = max{11, x1 + 5}
          add_A(x1,x2) = max{17, x2 + 7}
          if_min#_A(x1,x2) = max{18, x2 - 2}
          le_A(x1,x2) = max{x1, x2 + 54}
          true_A = 1
          |0|_A = 2
          s_A(x1) = max{58, x1}
          false_A = 57
    
      precedence:
      
        add = if_min# > min# = le = true > s = false > |0|
      
      partial status:
    
        pi(min#) = [1]
        pi(add) = [2]
        pi(if_min#) = []
        pi(le) = [1, 2]
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          min#_A(x1) = x1 + 5
          add_A(x1,x2) = x2
          if_min#_A(x1,x2) = 5
          le_A(x1,x2) = 7
          true_A = 8
          |0|_A = 7
          s_A(x1) = x1 + 12
          false_A = 11
    
      precedence:
      
        true = s = false > min# = add = if_min# = le = |0|
      
      partial status:
    
        pi(min#) = [1]
        pi(add) = [2]
        pi(if_min#) = []
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(false) = []
    

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: le#(s(x),s(y)) -> le#(x,y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

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:
        
          le#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = []
        pi(s) = [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: app#(add(n,x),y) -> app#(x,y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: le(|0|(),y) -> true()
r6: le(s(x),|0|()) -> false()
r7: le(s(x),s(y)) -> le(x,y)
r8: app(nil(),y) -> y
r9: app(add(n,x),y) -> add(n,app(x,y))
r10: min(add(n,nil())) -> n
r11: min(add(n,add(m,x))) -> if_min(le(n,m),add(n,add(m,x)))
r12: if_min(true(),add(n,add(m,x))) -> min(add(n,x))
r13: if_min(false(),add(n,add(m,x))) -> min(add(m,x))
r14: rm(n,nil()) -> nil()
r15: rm(n,add(m,x)) -> if_rm(eq(n,m),n,add(m,x))
r16: if_rm(true(),n,add(m,x)) -> rm(n,x)
r17: if_rm(false(),n,add(m,x)) -> add(m,rm(n,x))
r18: minsort(nil(),nil()) -> nil()
r19: minsort(add(n,x),y) -> if_minsort(eq(n,min(add(n,x))),add(n,x),y)
r20: if_minsort(true(),add(n,x),y) -> add(n,minsort(app(rm(n,x),y),nil()))
r21: if_minsort(false(),add(n,x),y) -> minsort(x,add(n,y))

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

The next rules are strictly ordered:

  p1

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