NO Termination w.r.t. Q proof of EEG_IJCAR_12_velroyen-nonloop-ConvLower_c.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 MNOCProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 NonLoopProof (⇒, 26 ms)
        ↳32 NO

(0) Obligation:

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

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

The TRS R 2 is

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))

The signature Sigma is {while}

(2) Obligation:

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

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

The set Q consists of the following terms:

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

(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:

WHILE(true, s(s(s(i)))) → WHILE(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
WHILE(true, s(s(s(i)))) → GT(s(s(s(i))), s(0))
WHILE(true, s(s(s(i)))) → F(s(s(s(i))))
F(i) → IF(neq(i, s(s(0))), i)
F(i) → NEQ(i, s(s(0)))
GT(s(x), s(y)) → GT(x, y)
IF(true, i) → PLUS(i, s(0))
NEQ(s(x), s(y)) → NEQ(x, y)
PLUS(s(x), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

The set Q consists of the following terms:

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

The set Q consists of the following terms:

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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:

PLUS(s(x), y) → PLUS(x, s(y))

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

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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].

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

(11) Obligation:

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

PLUS(s(x), y) → PLUS(x, s(y))

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:

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

(13) YES

(14) Obligation:

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

NEQ(s(x), s(y)) → NEQ(x, y)

The TRS R consists of the following rules:

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

The set Q consists of the following terms:

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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:

NEQ(s(x), s(y)) → NEQ(x, y)

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

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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].

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

(18) Obligation:

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

NEQ(s(x), s(y)) → NEQ(x, y)

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:

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

(20) YES

(21) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

The set Q consists of the following terms:

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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:

GT(s(x), s(y)) → GT(x, y)

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

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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].

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

(25) Obligation:

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

GT(s(x), s(y)) → GT(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:

  • GT(s(x), s(y)) → GT(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:

WHILE(true, s(s(s(i)))) → WHILE(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))

The TRS R consists of the following rules:

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

The set Q consists of the following terms:

while(true, s(s(s(x0))))
f(x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
gt(0, 0)
gt(0, s(x0))
if(true, x0)
if(false, x0)
neq(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
plus(s(x0), x1)
plus(0, x0)

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:

WHILE(true, s(s(s(i)))) → WHILE(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))

The TRS R consists of the following rules:

while(true, s(s(s(i)))) → while(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))
f(i) → if(neq(i, s(s(0))), i)
gt(s(x), s(y)) → gt(x, y)
gt(s(x), 0) → true
gt(0, 0) → false
gt(0, s(y)) → false
if(true, i) → plus(i, s(0))
if(false, i) → i
neq(s(x), s(y)) → neq(x, y)
neq(0, 0) → false
neq(0, s(y)) → true
neq(s(x), 0) → true
plus(s(x), y) → plus(x, s(y))
plus(0, y) → y

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
WHILE(true, s(s(s(s(zr0)))))[zr0 / s(zr0)]n[zr0 / 0] → WHILE(true, s(s(s(s(s(zr0))))))[zr0 / s(zr0)]n[zr0 / 0]
This rule is correct for the QDP as the following derivation shows:

WHILE(true, s(s(s(s(zr0)))))[zr0 / s(zr0)]n[zr0 / 0] → WHILE(true, s(s(s(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) - Equiv IPS (rhs) - Equiv IPS (lhs)
    WHILE(true, s(s(s(s(zl1)))))[zr1 / s(zr1), zl1 / s(zl1)]n[zr1 / 0, zl1 / 0] → WHILE(true, s(s(s(s(s(zr1))))))[zr1 / s(zr1), zl1 / s(zl1)]n[zr1 / 0, zl1 / 0]
        by Narrowing at position: [1]
        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 IPS (rhs) - Equiv IPS (lhs)
        WHILE(true, s(s(s(s(zs1)))))[zt1 / s(zt1), zs1 / s(zs1)]n[zt1 / 0, zs1 / y1] → WHILE(true, plus(y1, s(s(s(s(s(zt1)))))))[zt1 / s(zt1), zs1 / s(zs1)]n[zt1 / 0, zs1 / y1]
            by Narrowing at position: [1]
            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate Sigma - Instantiation - Instantiation
            WHILE(true, s(s(s(i))))[ ]n[ ] → WHILE(true, plus(i, s(s(s(s(0))))))[ ]n[ ]
                by Rewrite t with the rewrite sequence : [([0],gt(s(x), s(y)) -> gt(x, y)), ([0],gt(s(x), 0) -> true), ([1],f(i) -> if(neq(i, s(s(0))), i)), ([1,0],neq(s(x), s(y)) -> neq(x, y)), ([1,0],neq(s(x), s(y)) -> neq(x, y)), ([1,0],neq(s(x), 0) -> true), ([1],if(true, i) -> plus(i, s(0))), ([1],plus(s(x), y) -> plus(x, s(y))), ([1],plus(s(x), y) -> plus(x, s(y))), ([1],plus(s(x), y) -> plus(x, s(y)))]
                WHILE(true, s(s(s(i))))[ ]n[ ] → WHILE(gt(s(s(s(i))), s(0)), f(s(s(s(i)))))[ ]n[ ]
                    by Rule from TRS P

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

        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate Sigma - Instantiation - Instantiation
        plus(0, y)[ ]n[ ] → y[ ]n[ ]
            by Rule from TRS R

(32) NO