(0) Obligation:

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

f(tt, x) → f(eq(x, half(double(x))), s(x))
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → tt
double(s(x)) → s(s(double(x)))
double(0) → 0
half(s(s(x))) → s(half(x))
half(0) → 0

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:

F(tt, x) → F(eq(x, half(double(x))), s(x))
F(tt, x) → EQ(x, half(double(x)))
F(tt, x) → HALF(double(x))
F(tt, x) → DOUBLE(x)
EQ(s(x), s(y)) → EQ(x, y)
DOUBLE(s(x)) → DOUBLE(x)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

f(tt, x) → f(eq(x, half(double(x))), s(x))
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → tt
double(s(x)) → s(s(double(x)))
double(0) → 0
half(s(s(x))) → s(half(x))
half(0) → 0

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

f(tt, x) → f(eq(x, half(double(x))), s(x))
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → tt
double(s(x)) → s(s(double(x)))
double(0) → 0
half(s(s(x))) → s(half(x))
half(0) → 0

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

(6) 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:

  • HALF(s(s(x))) → HALF(x)
    The graph contains the following edges 1 > 1

(7) YES

(8) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

f(tt, x) → f(eq(x, half(double(x))), s(x))
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → tt
double(s(x)) → s(s(double(x)))
double(0) → 0
half(s(s(x))) → s(half(x))
half(0) → 0

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

(9) 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:

  • DOUBLE(s(x)) → DOUBLE(x)
    The graph contains the following edges 1 > 1

(10) YES

(11) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

f(tt, x) → f(eq(x, half(double(x))), s(x))
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → tt
double(s(x)) → s(s(double(x)))
double(0) → 0
half(s(s(x))) → s(half(x))
half(0) → 0

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

(12) 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:

  • EQ(s(x), s(y)) → EQ(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

F(tt, x) → F(eq(x, half(double(x))), s(x))

The TRS R consists of the following rules:

f(tt, x) → f(eq(x, half(double(x))), s(x))
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → tt
double(s(x)) → s(s(double(x)))
double(0) → 0
half(s(s(x))) → s(half(x))
half(0) → 0

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

(15) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 1,
σ' = [ ], and μ' = [ ] on the rule
F(tt, s(zr0))[zr0 / s(zr0)]n[zr0 / 0] → F(tt, s(s(zr0)))[zr0 / s(zr0)]n[zr0 / 0]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Domain Renaming) - Equivalent (Domain Renaming)
F(tt, s(zl1))[zl1 / s(zl1)]n[zl1 / 0] → F(tt, s(s(zr1)))[zr1 / s(zr1)]n[zr1 / 0]
    by Rewrite t
        intermediate steps: Equivalent (Remove Unused)
        F(tt, s(zl1))[zl1 / s(zl1), zr1 / s(zr1)]n[zl1 / 0, zr1 / 0] → F(eq(0, half(0)), s(s(zr1)))[zl1 / s(zl1), zr1 / s(zr1)]n[zl1 / 0, zr1 / 0]
            by Narrowing at position: [0,1,0]
                intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Simplify mu) - Equivalent (Remove Unused)
                F(tt, s(zl1))[zl1 / s(zl1), zr2 / s(zr2), zr3 / s(zr3)]n[zl1 / y1, x0 / y1, zr2 / y1, zr3 / half(double(y1)), y0 / half(double(y1))] → F(eq(y1, y0), s(s(zr2)))[zl1 / s(zl1), zr2 / s(zr2), zr3 / s(zr3)]n[zl1 / y1, x0 / y1, zr2 / y1, zr3 / half(double(y1)), y0 / half(double(y1))]
                    by Narrowing at position: [0]
                        intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
                        F(tt, s(zl1))[zl1 / s(zl1), zr2 / s(zr2), zr3 / s(s(zr3)), zt1 / s(zt1)]n[zl1 / x0, zr2 / x0, zr3 / double(x0), y0 / double(x0), zt1 / half(double(x0))] → F(eq(s(zr2), s(zt1)), s(s(zr2)))[zl1 / s(zl1), zr2 / s(zr2), zr3 / s(s(zr3)), zt1 / s(zt1)]n[zl1 / x0, zr2 / x0, zr3 / double(x0), y0 / double(x0), zt1 / half(double(x0))]
                            by Narrowing at position: [0,1]
                                intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
                                F(tt, s(zs1))[zs1 / s(zs1), zt1 / s(s(zt1))]n[zs1 / y0, zt1 / double(y0)] → F(eq(s(zs1), half(s(s(zt1)))), s(s(zs1)))[zs1 / s(zs1), zt1 / s(s(zt1))]n[zs1 / y0, zt1 / double(y0)]
                                    by Narrowing at position: [0,1,0]
                                        intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Instantiate Sigma - Instantiation - Instantiation
                                        F(tt, x)[ ]n[ ] → F(eq(x, half(double(x))), s(x))[ ]n[ ]
                                            by OriginalRule from TRS P

                                        intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
                                        double(s(x))[x / s(x)]n[ ] → s(s(z))[x / s(x), z / s(s(z))]n[z / double(x)]
                                            by PatternCreation II
                                                double(s(x))[ ]n[ ] → s(s(double(x)))[ ]n[ ]
                                                    by OriginalRule from TRS R

                                intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
                                half(s(s(x)))[x / s(s(x))]n[ ] → s(z)[x / s(s(x)), z / s(z)]n[z / half(x)]
                                    by PatternCreation II
                                        half(s(s(x)))[ ]n[ ] → s(half(x))[ ]n[ ]
                                            by OriginalRule from TRS R

                        intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
                        eq(s(x), s(y))[x / s(x), y / s(y)]n[ ] → eq(x, y)[ ]n[ ]
                            by PatternCreation I
                                eq(s(x), s(y))[ ]n[ ] → eq(x, y)[ ]n[ ]
                                    by OriginalRule from TRS R

                intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
                double(0)[ ]n[ ] → 0[ ]n[ ]
                    by OriginalRule from TRS R

(16) NO