YES Termination w.r.t. Q proof of Zantema_05_z10.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔, 14 ms)
↳2 QDP
↳3 QDPOrderProof (⇔, 15 ms)
↳4 QDP
↳5 QDPSizeChangeProof (⇔, 0 ms)
↳6 YES

(0) Obligation:

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

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, p(x, y)) → y

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

A(lambda(x), y) → A(x, p(1, a(y, t)))
A(lambda(x), y) → A(y, t)
A(p(x, y), z) → A(x, z)
A(p(x, y), z) → A(y, z)
A(a(x, y), z) → A(x, a(y, z))
A(a(x, y), z) → A(y, z)

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, p(x, y)) → y

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


A(lambda(x), y) → A(x, p(1, a(y, t)))
A(lambda(x), y) → A(y, t)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation with max and min functions [POLO,MAXPOLO]:

POL(1) = 0   
POL(A(x1, x2)) = x1 + x2   
POL(a(x1, x2)) = x1 + x2   
POL(id) = 0   
POL(lambda(x1)) = 1 + x1   
POL(p(x1, x2)) = max(x1, x2)   
POL(t) = 0   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, p(x, y)) → y

(4) Obligation:

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

A(p(x, y), z) → A(x, z)
A(p(x, y), z) → A(y, z)
A(a(x, y), z) → A(x, a(y, z))
A(a(x, y), z) → A(y, z)

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, p(x, y)) → y

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

(5) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • A(p(x, y), z) → A(x, z)
    The graph contains the following edges 1 > 1, 2 >= 2

  • A(p(x, y), z) → A(y, z)
    The graph contains the following edges 1 > 1, 2 >= 2

  • A(a(x, y), z) → A(x, a(y, z))
    The graph contains the following edges 1 > 1

  • A(a(x, y), z) → A(y, z)
    The graph contains the following edges 1 > 1, 2 >= 2

(6) YES