YES Termination proof of ../tpdb/TRS/CSR/Ex6_9_Luc02c.trs
Termination of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(X)))

The replacement map contains the following entries:

2nd: {1}
cons1: {1, 2}
cons: {1}
from: {1}
s: {1}


↳ CSR
  ↳ Zantema-Transformation

Q restricted rewrite system:
The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(X)))

The replacement map contains the following entries:

2nd: {1}
cons1: {1, 2}
cons: {1}
from: {1}
s: {1}

We applied the Zantema [34] to transform the context-sensitive TRS to an usual TRS.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, a(X1)))
from(X) → cons(X, fromInact(s(X)))
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A(fromInact(x1)) → FROM(x1)
2ND(cons(X, X1)) → 2ND(cons1(X, a(X1)))
2ND(cons(X, X1)) → A(X1)

The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, a(X1)))
from(X) → cons(X, fromInact(s(X)))
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

A(fromInact(x1)) → FROM(x1)
2ND(cons(X, X1)) → 2ND(cons1(X, a(X1)))
2ND(cons(X, X1)) → A(X1)

The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, a(X1)))
from(X) → cons(X, fromInact(s(X)))
a(x) → x
from(x1) → fromInact(x1)
a(fromInact(x1)) → from(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.