YES

We show the termination of the TRS R:

  first(|0|(),X) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
  from(X) -> cons(X,n__from(s(X)))
  first(X1,X2) -> n__first(X1,X2)
  from(X) -> n__from(X)
  activate(n__first(X1,X2)) -> first(X1,X2)
  activate(n__from(X)) -> from(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__first(X1,X2)) -> first#(X1,X2)
p3: activate#(n__from(X)) -> from#(X)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: activate(n__first(X1,X2)) -> first(X1,X2)
r7: activate(n__from(X)) -> from(X)
r8: activate(X) -> 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: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: activate(n__first(X1,X2)) -> first(X1,X2)
r7: activate(n__from(X)) -> from(X)
r8: activate(X) -> 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:
      
        first#_A(x1,x2) = max{5, x1 - 1, x2 + 3}
        s_A(x1) = x1
        cons_A(x1,x2) = max{x1 - 1, x2 + 3}
        activate#_A(x1) = x1 + 5
        n__first_A(x1,x2) = max{x1 - 6, x2}
  
    precedence:
    
      first# = s = cons = activate# = n__first
    
    partial status:
  
      pi(first#) = []
      pi(s) = [1]
      pi(cons) = [2]
      pi(activate#) = [1]
      pi(n__first) = [2]

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: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: activate(n__first(X1,X2)) -> first(X1,X2)
r7: activate(n__from(X)) -> from(X)
r8: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)