(0) Obligation:

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(f(x1, x2, x3)) = x1 + x2 + 2·x3   
POL(g(x1, x2)) = 2 + x1 + x2   
POL(h(x1, x2)) = 2·x1 + 2·x2   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

g(0, Y) → 0


(2) Obligation:

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, 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:

H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))
G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(6) Complex Obligation (AND)

(7) Obligation:

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

G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

(8) 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:

  • G(X, s(Y)) → G(X, Y)
    The graph contains the following edges 1 >= 1, 2 > 2

(9) YES

(10) Obligation:

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

F(X, Y, g(X, Y)) → H(0, g(X, Y))
H(X, Z) → F(X, s(X), Z)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

(11) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [ ] on the rule
H(0, g(0, s(0)))[ ]n[x0 / 0] → H(0, g(0, s(0)))[ ]n[x0 / 0]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
H(x1, g(x1, s(x1)))[ ]n[ ] → H(0, g(x1, s(x1)))[ ]n[ ]
    by Narrowing at position: []
        intermediate steps: Instantiation - Instantiation
        H(X, Z)[ ]n[ ] → F(X, s(X), Z)[ ]n[ ]
            by OriginalRule from TRS P

        intermediate steps: Instantiation - Instantiation - Instantiation
        F(X, Y, g(X, Y))[ ]n[ ] → H(0, g(X, Y))[ ]n[ ]
            by OriginalRule from TRS P

(12) NO