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

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

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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