YES

We show the termination of the TRS R:

  a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
  a__from(X) -> cons(mark(X),from(s(X)))
  mark(|2nd|(X)) -> a__2nd(mark(X))
  mark(from(X)) -> a__from(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  a__2nd(X) -> |2nd|(X)
  a__from(X) -> from(X)

-- SCC decomposition.

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

p1: a__2nd#(cons(X,cons(Y,Z))) -> mark#(Y)
p2: a__from#(X) -> mark#(X)
p3: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p4: mark#(|2nd|(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: a__2nd#(cons(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(|2nd|(X)) -> mark#(X)
p8: mark#(|2nd|(X)) -> a__2nd#(mark(X))

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    a__2nd#_A(x1) = x1 + 8
    cons_A(x1,x2) = max{3, x1, x2 - 3}
    mark#_A(x1) = max{1, x1}
    s_A(x1) = max{3, x1}
    from_A(x1) = max{7, x1 + 6}
    a__from#_A(x1) = max{1, x1}
    mark_A(x1) = x1 + 6
    |2nd|_A(x1) = max{15, x1 + 14}
    a__2nd_A(x1) = max{15, x1 + 14}
    a__from_A(x1) = max{7, x1 + 6}

The next rules are strictly ordered:

  p1, p4, p7

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> a__from#(mark(X))
p4: a__from#(X) -> mark#(X)
p5: mark#(|2nd|(X)) -> a__2nd#(mark(X))

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(from(X)) -> a__from#(mark(X))
p3: a__from#(X) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = max{0, x1 - 2}
    s_A(x1) = x1 + 1
    from_A(x1) = max{6, x1 + 5}
    a__from#_A(x1) = x1 + 1
    mark_A(x1) = x1 + 2
    cons_A(x1,x2) = max{x1 + 2, x2 - 1}
    a__2nd_A(x1) = max{3, x1 + 2}
    a__from_A(x1) = max{6, x1 + 5}
    |2nd|_A(x1) = max{3, x1 + 2}

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(from(X)) -> a__from#(mark(X))
p3: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p3}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = max{0, x1 - 1}
    s_A(x1) = x1 + 1
    cons_A(x1,x2) = max{x1 + 2, x2 + 2}

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: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = x1
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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