YES

We show the termination of the TRS R:

  eq(|0|(),|0|()) -> true()
  eq(|0|(),s(m)) -> false()
  eq(s(n),|0|()) -> false()
  eq(s(n),s(m)) -> eq(n,m)
  le(|0|(),m) -> true()
  le(s(n),|0|()) -> false()
  le(s(n),s(m)) -> le(n,m)
  min(cons(|0|(),nil())) -> |0|()
  min(cons(s(n),nil())) -> s(n)
  min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
  if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
  if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
  replace(n,m,nil()) -> nil()
  replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
  if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
  if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
  |sort|(nil()) -> nil()
  |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

-- SCC decomposition.

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

p1: eq#(s(n),s(m)) -> eq#(n,m)
p2: le#(s(n),s(m)) -> le#(n,m)
p3: min#(cons(n,cons(m,x))) -> if_min#(le(n,m),cons(n,cons(m,x)))
p4: min#(cons(n,cons(m,x))) -> le#(n,m)
p5: if_min#(true(),cons(n,cons(m,x))) -> min#(cons(n,x))
p6: if_min#(false(),cons(n,cons(m,x))) -> min#(cons(m,x))
p7: replace#(n,m,cons(k,x)) -> if_replace#(eq(n,k),n,m,cons(k,x))
p8: replace#(n,m,cons(k,x)) -> eq#(n,k)
p9: if_replace#(false(),n,m,cons(k,x)) -> replace#(n,m,x)
p10: |sort|#(cons(n,x)) -> min#(cons(n,x))
p11: |sort|#(cons(n,x)) -> |sort|#(replace(min(cons(n,x)),n,x))
p12: |sort|#(cons(n,x)) -> replace#(min(cons(n,x)),n,x)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The estimated dependency graph contains the following SCCs:

  {p11}
  {p7, p9}
  {p1}
  {p3, p5, p6}
  {p2}


-- Reduction pair.

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

p1: |sort|#(cons(n,x)) -> |sort|#(replace(min(cons(n,x)),n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          |sort|#_A(x1) = x1 + (3,8)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (10,8)
          replace_A(x1,x2,x3) = x2 + ((1,0),(1,0)) x3 + (1,1)
          min_A(x1) = ((1,0),(0,0)) x1 + (2,10)
          eq_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,3)
          |0|_A() = (0,7)
          true_A() = (0,6)
          s_A(x1) = ((1,0),(0,0)) x1 + (12,4)
          false_A() = (0,0)
          le_A(x1,x2) = ((0,0),(1,0)) x2 + (11,5)
          if_min_A(x1,x2) = x2 + (1,1)
          if_replace_A(x1,x2,x3,x4) = ((0,0),(1,0)) x1 + x3 + ((1,0),(0,0)) x4 + (1,9)
          nil_A() = (1,3)
  
    precedence:
    
      replace = eq = true = le = if_replace = nil > |sort|# > |0| > cons = min = s > false > if_min
    
    partial status:
  
      pi(|sort|#) = [1]
      pi(cons) = []
      pi(replace) = []
      pi(min) = []
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []
      pi(false) = []
      pi(le) = []
      pi(if_min) = [2]
      pi(if_replace) = []
      pi(nil) = []

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_replace#(false(),n,m,cons(k,x)) -> replace#(n,m,x)
p2: replace#(n,m,cons(k,x)) -> if_replace#(eq(n,k),n,m,cons(k,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          if_replace#_A(x1,x2,x3,x4) = x2 + x4 + (1,2)
          false_A() = (1,3)
          cons_A(x1,x2) = ((1,0),(1,1)) x2 + (3,1)
          replace#_A(x1,x2,x3) = x1 + ((1,0),(1,1)) x3 + (2,2)
          eq_A(x1,x2) = x1 + (4,0)
          |0|_A() = (2,1)
          true_A() = (3,0)
          s_A(x1) = x1 + (0,4)
  
    precedence:
    
      if_replace# = false = replace# > eq > true > cons = |0| = s
    
    partial status:
  
      pi(if_replace#) = [4]
      pi(false) = []
      pi(cons) = []
      pi(replace#) = [3]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []

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: replace#(n,m,cons(k,x)) -> if_replace#(eq(n,k),n,m,cons(k,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

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(n),s(m)) -> eq#(n,m)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          eq#_A(x1,x2) = x1
          s_A(x1) = x1 + (1,1)
  
    precedence:
    
      s > eq#
    
    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(),cons(n,cons(m,x))) -> min#(cons(m,x))
p2: min#(cons(n,cons(m,x))) -> if_min#(le(n,m),cons(n,cons(m,x)))
p3: if_min#(true(),cons(n,cons(m,x))) -> min#(cons(n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          if_min#_A(x1,x2) = x2 + (1,1)
          false_A() = (0,0)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (3,3)
          min#_A(x1) = ((1,0),(0,0)) x1 + (2,4)
          le_A(x1,x2) = ((0,0),(1,0)) x2 + (4,5)
          true_A() = (1,6)
          |0|_A() = (2,7)
          s_A(x1) = ((1,0),(0,0)) x1 + (5,1)
  
    precedence:
    
      min# > cons > if_min# = false = le = true = |0| = s
    
    partial status:
  
      pi(if_min#) = [2]
      pi(false) = []
      pi(cons) = [2]
      pi(min#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = []

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

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

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

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          min#_A(x1) = ((0,0),(1,0)) x1 + (1,2)
          cons_A(x1,x2) = x2 + (3,9)
          if_min#_A(x1,x2) = ((0,0),(1,0)) x2 + (1,1)
          le_A(x1,x2) = ((1,0),(1,0)) x1 + (3,7)
          true_A() = (2,6)
          |0|_A() = (3,7)
          s_A(x1) = ((1,0),(0,0)) x1 + (4,8)
          false_A() = (0,0)
  
    precedence:
    
      cons > min# = s = false > le > true = |0| > if_min#
    
    partial status:
  
      pi(min#) = []
      pi(cons) = [2]
      pi(if_min#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = []
      pi(false) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: min#(cons(n,cons(m,x))) -> if_min#(le(n,m),cons(n,cons(m,x)))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: le#(s(n),s(m)) -> le#(n,m)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          le#_A(x1,x2) = x1
          s_A(x1) = x1 + (1,1)
  
    precedence:
    
      s > le#
    
    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.