YES

We show the termination of the TRS R:

  a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
  a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
  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))
  mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
  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#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))
p3: a__2nd#(cons(X,X1)) -> mark#(X)
p4: a__2nd#(cons(X,X1)) -> mark#(X1)
p5: a__from#(X) -> mark#(X)
p6: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p7: mark#(|2nd|(X)) -> mark#(X)
p8: mark#(from(X)) -> a__from#(mark(X))
p9: mark#(from(X)) -> mark#(X)
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(s(X)) -> mark#(X)
p12: mark#(cons1(X1,X2)) -> mark#(X1)
p13: mark#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13}


-- Reduction pair.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(cons1(X1,X2)) -> mark#(X2)
p3: mark#(cons1(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: a__from#(X) -> mark#(X)
p9: mark#(|2nd|(X)) -> mark#(X)
p10: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p11: a__2nd#(cons(X,X1)) -> mark#(X1)
p12: a__2nd#(cons(X,X1)) -> mark#(X)
p13: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      a__2nd#_A(x1) = x1 + 2
      cons1_A(x1,x2) = max{3, x1, x2}
      cons_A(x1,x2) = max{4, x1 + 2, x2}
      mark#_A(x1) = max{1, x1 - 3}
      s_A(x1) = x1
      from_A(x1) = x1 + 5
      a__from#_A(x1) = x1 + 1
      mark_A(x1) = x1
      |2nd|_A(x1) = x1 + 6
      a__2nd_A(x1) = x1 + 6
      a__from_A(x1) = x1 + 5
    
    2. max/plus interpretations on natural numbers:
    
      a__2nd#_A(x1) = max{3, x1 + 2}
      cons1_A(x1,x2) = 0
      cons_A(x1,x2) = 3
      mark#_A(x1) = 1
      s_A(x1) = x1
      from_A(x1) = x1
      a__from#_A(x1) = x1 + 1
      mark_A(x1) = 5
      |2nd|_A(x1) = max{1, x1 - 1}
      a__2nd_A(x1) = max{1, x1 - 1}
      a__from_A(x1) = max{4, x1}
    

The next rules are strictly ordered:

  p1, p6, p7, p9, p10, p11, p12, p13

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(cons1(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: a__from#(X) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: 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#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons1(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      mark#_A(x1) = max{0, x1 - 1}
      cons1_A(x1,x2) = max{x1 + 2, x2 + 2}
      cons_A(x1,x2) = max{x1, x2 + 1}
      s_A(x1) = x1 + 2
    
    2. max/plus interpretations on natural numbers:
    
      mark#_A(x1) = 0
      cons1_A(x1,x2) = 0
      cons_A(x1,x2) = 0
      s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p3, p4

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: 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#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      mark#_A(x1) = max{1, x1}
      cons_A(x1,x2) = max{x1 + 2, x2 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      mark#_A(x1) = 0
      cons_A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1

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