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)
  rm(N,nil()) -> nil()
  rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
  ifrm(true(),N,add(M,X)) -> rm(N,X)
  ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
  purge(nil()) -> nil()
  purge(add(N,X)) -> add(N,purge(rm(N,X)))

-- SCC decomposition.

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

p1: eq#(s(X),s(Y)) -> eq#(X,Y)
p2: rm#(N,add(M,X)) -> ifrm#(eq(N,M),N,add(M,X))
p3: rm#(N,add(M,X)) -> eq#(N,M)
p4: ifrm#(true(),N,add(M,X)) -> rm#(N,X)
p5: ifrm#(false(),N,add(M,X)) -> rm#(N,X)
p6: purge#(add(N,X)) -> purge#(rm(N,X))
p7: purge#(add(N,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: rm(N,nil()) -> nil()
r6: rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
r7: ifrm(true(),N,add(M,X)) -> rm(N,X)
r8: ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
r9: purge(nil()) -> nil()
r10: purge(add(N,X)) -> add(N,purge(rm(N,X)))

The estimated dependency graph contains the following SCCs:

  {p6}
  {p2, p4, p5}
  {p1}


-- Reduction pair.

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

p1: purge#(add(N,X)) -> purge#(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: rm(N,nil()) -> nil()
r6: rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
r7: ifrm(true(),N,add(M,X)) -> rm(N,X)
r8: ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
r9: purge(nil()) -> nil()
r10: purge(add(N,X)) -> add(N,purge(rm(N,X)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          purge#_A(x1) = x1 + (1,0)
          add_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,1),(1,1)) x2 + (4,4)
          rm_A(x1,x2) = ((1,1),(1,1)) x2 + (2,2)
          eq_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,0),(1,0)) x2 + (11,0)
          |0|_A() = (2,1)
          true_A() = (1,1)
          s_A(x1) = x1 + (2,1)
          false_A() = (1,0)
          ifrm_A(x1,x2,x3) = ((1,1),(1,1)) x3 + (1,0)
          nil_A() = (1,1)
  
    precedence:
    
      purge# > add = rm = s = false = ifrm > nil > eq > true > |0|
    
    partial status:
  
      pi(purge#) = [1]
      pi(add) = []
      pi(rm) = []
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(false) = []
      pi(ifrm) = []
      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: ifrm#(false(),N,add(M,X)) -> rm#(N,X)
p2: rm#(N,add(M,X)) -> ifrm#(eq(N,M),N,add(M,X))
p3: ifrm#(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: rm(N,nil()) -> nil()
r6: rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
r7: ifrm(true(),N,add(M,X)) -> rm(N,X)
r8: ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
r9: purge(nil()) -> nil()
r10: purge(add(N,X)) -> add(N,purge(rm(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: standard order
        interpretations:
          ifrm#_A(x1,x2,x3) = ((1,1),(0,0)) x3 + (1,3)
          false_A() = (3,2)
          add_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,1)) x2 + (5,4)
          rm#_A(x1,x2) = ((1,1),(0,0)) x2 + (4,3)
          eq_A(x1,x2) = (14,5)
          true_A() = (1,2)
          |0|_A() = (2,1)
          s_A(x1) = (15,5)
  
    precedence:
    
      s > eq = |0| > ifrm# = rm# > add = true > false
    
    partial status:
  
      pi(ifrm#) = []
      pi(false) = []
      pi(add) = [2]
      pi(rm#) = []
      pi(eq) = []
      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: rm#(N,add(M,X)) -> ifrm#(eq(N,M),N,add(M,X))
p2: ifrm#(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: rm(N,nil()) -> nil()
r6: rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
r7: ifrm(true(),N,add(M,X)) -> rm(N,X)
r8: ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
r9: purge(nil()) -> nil()
r10: purge(add(N,X)) -> add(N,purge(rm(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: rm#(N,add(M,X)) -> ifrm#(eq(N,M),N,add(M,X))
p2: ifrm#(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: rm(N,nil()) -> nil()
r6: rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
r7: ifrm(true(),N,add(M,X)) -> rm(N,X)
r8: ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
r9: purge(nil()) -> nil()
r10: purge(add(N,X)) -> add(N,purge(rm(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: standard order
        interpretations:
          rm#_A(x1,x2) = ((0,1),(0,0)) x2 + (3,2)
          add_A(x1,x2) = ((1,1),(0,1)) x2 + (5,3)
          ifrm#_A(x1,x2,x3) = ((0,1),(0,0)) x3 + (1,2)
          eq_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (7,4)
          true_A() = (1,1)
          |0|_A() = (3,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          false_A() = (2,2)
  
    precedence:
    
      true > eq > ifrm# = |0| > s = false > rm# = add
    
    partial status:
  
      pi(rm#) = []
      pi(add) = []
      pi(ifrm#) = []
      pi(eq) = []
      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: rm#(N,add(M,X)) -> ifrm#(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: rm(N,nil()) -> nil()
r6: rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
r7: ifrm(true(),N,add(M,X)) -> rm(N,X)
r8: ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
r9: purge(nil()) -> nil()
r10: purge(add(N,X)) -> add(N,purge(rm(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(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: rm(N,nil()) -> nil()
r6: rm(N,add(M,X)) -> ifrm(eq(N,M),N,add(M,X))
r7: ifrm(true(),N,add(M,X)) -> rm(N,X)
r8: ifrm(false(),N,add(M,X)) -> add(M,rm(N,X))
r9: purge(nil()) -> nil()
r10: purge(add(N,X)) -> add(N,purge(rm(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: standard order
        interpretations:
          eq#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (2,2)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      eq# = s
    
    partial status:
  
      pi(eq#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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