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:
  
      max/plus interpretations on natural numbers:
      
        purge#_A(x1) = x1 + 1
        add_A(x1,x2) = max{x1 + 4, x2 + 4}
        rm_A(x1,x2) = x2 + 2
        eq_A(x1,x2) = x2 + 7
        |0|_A = 10
        true_A = 11
        s_A(x1) = x1 + 12
        false_A = 9
        ifrm_A(x1,x2,x3) = max{6, x1 - 1, x3 + 2}
        nil_A = 2
  
    precedence:
    
      rm = ifrm > add > purge# > eq = |0| = true = s = false = nil
    
    partial status:
  
      pi(purge#) = [1]
      pi(add) = [2]
      pi(rm) = []
      pi(eq) = [2]
      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:
  
      max/plus interpretations on natural numbers:
      
        ifrm#_A(x1,x2,x3) = max{x1 - 3, x3 + 1}
        false_A = 11
        add_A(x1,x2) = max{x1 - 3, x2 + 5}
        rm#_A(x1,x2) = x2 + 2
        eq_A(x1,x2) = max{9, x2 + 1}
        true_A = 6
        |0|_A = 23
        s_A(x1) = x1 + 10
  
    precedence:
    
      ifrm# = false = add = rm# = eq = true > |0| = s
    
    partial status:
  
      pi(ifrm#) = [3]
      pi(false) = []
      pi(add) = [2]
      pi(rm#) = []
      pi(eq) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

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:
  
      max/plus interpretations on natural numbers:
      
        rm#_A(x1,x2) = max{x1 + 7, x2 + 3}
        add_A(x1,x2) = x2 + 4
        ifrm#_A(x1,x2,x3) = max{x2 + 7, x3}
        eq_A(x1,x2) = max{8, x1 - 3}
        true_A = 1
        |0|_A = 0
        s_A(x1) = max{12, x1 + 2}
        false_A = 1
  
    precedence:
    
      rm# = add = ifrm# = eq = true = |0| = s = false
    
    partial status:
  
      pi(rm#) = [2]
      pi(add) = [2]
      pi(ifrm#) = [3]
      pi(eq) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      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:
  
      max/plus interpretations on natural numbers:
      
        eq#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 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.