YES Termination w.r.t. Q proof of Zantema_05_z29.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 MRRProof (⇔, 0 ms)
↳6 QDP
↳7 QDPSizeChangeProof (⇔, 0 ms)
↳8 YES

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(1) = 0   
POL(a(x1, x2)) = x1 + x2   
POL(lambda(x1)) = 2 + x1   
POL(t) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

lambda(x) → x


(2) Obligation:

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

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

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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

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

(5) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: Polynomial interpretation [POLO]:

POL(1) = 0   
POL(A(x1, x2)) = 2·x1 + 2·x2   
POL(a(x1, x2)) = x1 + x2   
POL(lambda(x1)) = 2 + x1   
POL(t) = 0   

(6) Obligation:

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

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

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

(7) 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(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

(8) YES