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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          purge#_A(x1) = max{6, x1 + 4}
          add_A(x1,x2) = max{x1 + 18, x2 + 12}
          rm_A(x1,x2) = max{7, x2 + 2}
          eq_A(x1,x2) = max{x1 + 4, x2 + 4}
          |0|_A = 1
          true_A = 0
          s_A(x1) = max{1, x1}
          false_A = 2
          ifrm_A(x1,x2,x3) = max{5, x3 + 2}
          nil_A = 14
    
      precedence:
      
        s > purge# = eq = false = nil > |0| > true > rm = ifrm > add
      
      partial status:
    
        pi(purge#) = [1]
        pi(add) = []
        pi(rm) = [2]
        pi(eq) = [1, 2]
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
        pi(false) = []
        pi(ifrm) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          purge#_A(x1) = max{1, x1}
          add_A(x1,x2) = 0
          rm_A(x1,x2) = 2
          eq_A(x1,x2) = max{x1 + 6, x2 - 9}
          |0|_A = 17
          true_A = 18
          s_A(x1) = x1 + 8
          false_A = 7
          ifrm_A(x1,x2,x3) = 2
          nil_A = 3
    
      precedence:
      
        purge# = add = rm = |0| = true = s = false = ifrm > eq = nil
      
      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:

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

  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.