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. lexicographic path order with precedence:
    
      precedence:
      
        eq > true > nil > add > rm > ifrm > false > s > |0| > purge#
      
      argument filter:
    
        pi(purge#) = [1]
        pi(add) = 2
        pi(rm) = 2
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = []
        pi(false) = []
        pi(ifrm) = 3
        pi(nil) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        nil > rm > ifrm > add > false > eq > s > true > |0| > purge#
      
      argument filter:
    
        pi(purge#) = [1]
        pi(add) = 2
        pi(rm) = 2
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = []
        pi(false) = []
        pi(ifrm) = 3
        pi(nil) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        nil > true > |0| > purge# > add > rm > ifrm > false > eq > s
      
      argument filter:
    
        pi(purge#) = [1]
        pi(add) = [2]
        pi(rm) = 2
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = []
        pi(false) = []
        pi(ifrm) = 3
        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. lexicographic path order with precedence:
    
      precedence:
      
        eq > false > s > true > |0| > ifrm# > add > rm#
      
      argument filter:
    
        pi(ifrm#) = 3
        pi(false) = []
        pi(add) = [1, 2]
        pi(rm#) = 2
        pi(eq) = [1, 2]
        pi(true) = []
        pi(|0|) = []
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        false > eq > s > true > |0| > rm# > ifrm# > add
      
      argument filter:
    
        pi(ifrm#) = 3
        pi(false) = []
        pi(add) = 1
        pi(rm#) = 2
        pi(eq) = 2
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        false > eq > s > true > |0| > rm# > ifrm# > add
      
      argument filter:
    
        pi(ifrm#) = 3
        pi(false) = []
        pi(add) = []
        pi(rm#) = []
        pi(eq) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = 1
    

The next rules are strictly ordered:

  p1, p2, p3

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: 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. lexicographic path order with precedence:
    
      precedence:
      
        s > eq#
      
      argument filter:
    
        pi(eq#) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > eq#
      
      argument filter:
    
        pi(eq#) = 1
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > eq#
      
      argument filter:
    
        pi(eq#) = 1
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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