NO Termination w.r.t. Q proof of AProVE_10_halfdouble.ari

(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) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

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

The TRS R 2 is

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

The signature Sigma is {f}

(2) 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

The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

(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:

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

The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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

The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

HALF(s(s(x))) → HALF(x)

R is empty.
The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

(11) Obligation:

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

HALF(s(s(x))) → HALF(x)

R is empty.
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:

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

(13) YES

(14) 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

The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

R is empty.
The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

(18) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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

(19) 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

(20) YES

(21) 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

The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

(25) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

(26) 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

(27) YES

(28) 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

The set Q consists of the following terms:

f(tt, x0)
eq(s(x0), s(x1))
eq(0, 0)
double(s(x0))
double(0)
half(s(s(x0)))
half(0)

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

(29) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(30) 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 (P,Q,R)-chains.

(31) NonLoopProof (COMPLETE 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:

F(tt, s(zr0))[zr0 / s(zr0)]n[zr0 / 0] → F(tt, s(s(zr0)))[zr0 / s(zr0)]n[zr0 / 0]
    by Equivalence by Domain Renaming of the lhs with [zl0 / zr0]
    intermediate steps: Equiv DR (rhs) - Equiv DR (lhs)
    F(tt, s(zl1))[zl1 / s(zl1)]n[zl1 / 0] → F(tt, s(s(zr1)))[zr1 / s(zr1)]n[zr1 / 0]
        by Rewrite t with the rewrite sequence : [([0,1],half(0) -> 0), ([0],eq(0, 0) -> tt)]
        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs)
        F(tt, s(zl1))[zr1 / s(zr1), zl1 / s(zl1)]n[zr1 / 0, zl1 / 0] → F(eq(0, half(0)), s(s(zr1)))[zr1 / s(zr1), zl1 / s(zl1)]n[zr1 / 0, zl1 / 0]
            by Narrowing at position: [0,1,0]
            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv Sµ (rhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
            F(tt, s(zl1))[zr2 / s(zr2), zr3 / s(zr3), zl1 / s(zl1)]n[zr2 / y1, zr3 / half(double(y1)), y0 / half(double(y1)), zl1 / y1, x0 / y1] → F(eq(y1, y0), s(s(zr2)))[zr2 / s(zr2), zr3 / s(zr3), zl1 / s(zl1)]n[zr2 / y1, zr3 / half(double(y1)), y0 / half(double(y1)), zl1 / y1, x0 / y1]
                by Narrowing at position: [0]
                intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
                F(tt, s(zl1))[zr2 / s(s(zr2)), zr3 / s(zr3), zt1 / s(zt1), zl1 / s(zl1)]n[zr2 / double(x0), zr3 / x0, y0 / double(x0), zt1 / half(double(x0)), zl1 / x0] → F(eq(s(zr3), s(zt1)), s(s(zr3)))[zr2 / s(s(zr2)), zr3 / s(zr3), zt1 / s(zt1), zl1 / s(zl1)]n[zr2 / double(x0), zr3 / x0, y0 / double(x0), zt1 / half(double(x0)), zl1 / x0]
                    by Narrowing at position: [0,1]
                    intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (lhs)
                    F(tt, s(zs1))[zt1 / s(s(zt1)), zs1 / s(zs1)]n[zt1 / double(y0), zs1 / y0] → F(eq(s(zs1), half(s(s(zt1)))), s(s(zs1)))[zt1 / s(s(zt1)), zs1 / s(zs1)]n[zt1 / double(y0), zs1 / y0]
                        by Narrowing at position: [0,1,0]
                        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate Sigma - Instantiation - Instantiation
                        F(tt, x)[ ]n[ ] → F(eq(x, half(double(x))), s(x))[ ]n[ ]
                            by Rule from TRS P

                        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs)
                        double(s(x))[x / s(x)]n[ ] → s(s(z))[x / s(x), z / s(s(z))]n[z / double(x)]
                            by PatternCreation II with pi: [0,0], sigma: [x / s(x)]
                            double(s(x))[ ]n[ ] → s(s(double(x)))[ ]n[ ]
                                by Rule from TRS R

                    intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (lhs) - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs)
                    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 with pi: [0], sigma: [x / s(s(x))]
                        half(s(s(x)))[ ]n[ ] → s(half(x))[ ]n[ ]
                            by Rule from TRS R

                intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv DR (lhs) - Equiv DR (lhs) - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
                eq(s(x), s(y))[x / s(x), y / s(y)]n[ ] → eq(x, y)[ ]n[ ]
                    by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x), y / s(y)]
                    eq(s(x), s(y))[ ]n[ ] → eq(x, y)[ ]n[ ]
                        by Rule from TRS R

            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
            double(0)[ ]n[ ] → 0[ ]n[ ]
                by Rule from TRS R

(32) NO