YES Termination w.r.t. Q proof of AProVE_04_Liveness8.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QDPOrderProof (⇔, 14 ms)
        ↳34 QDP
        ↳35 MRRProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 UsableRulesProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 MRRProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 PisEmptyProof (⇔, 0 ms)
        ↳42 YES

(0) Obligation:

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

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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:

TOP(active(c)) → TOP(mark(c))
TOP(mark(x)) → TOP(check(x))
TOP(mark(x)) → CHECK(x)
CHECK(f(x)) → F(check(x))
CHECK(f(x)) → CHECK(x)
CHECK(x) → START(match(f(X), x))
CHECK(x) → MATCH(f(X), x)
CHECK(x) → F(X)
MATCH(f(x), f(y)) → F(match(x, y))
MATCH(f(x), f(y)) → MATCH(x, y)
MATCH(X, x) → PROPER(x)
PROPER(f(x)) → F(proper(x))
PROPER(f(x)) → PROPER(x)
F(ok(x)) → F(x)
F(found(x)) → F(x)
TOP(found(x)) → TOP(active(x))
TOP(found(x)) → ACTIVE(x)
ACTIVE(f(x)) → F(active(x))
ACTIVE(f(x)) → ACTIVE(x)
F(mark(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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 6 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(found(x)) → F(x)
F(ok(x)) → F(x)
F(mark(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

F(found(x)) → F(x)
F(ok(x)) → F(x)
F(mark(x)) → F(x)

R is empty.
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:

  • F(found(x)) → F(x)
    The graph contains the following edges 1 > 1

  • F(ok(x)) → F(x)
    The graph contains the following edges 1 > 1

  • F(mark(x)) → F(x)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

ACTIVE(f(x)) → ACTIVE(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

ACTIVE(f(x)) → ACTIVE(x)

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

(13) 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:

  • ACTIVE(f(x)) → ACTIVE(x)
    The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

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

PROPER(f(x)) → PROPER(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

PROPER(f(x)) → PROPER(x)

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

(18) 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:

  • PROPER(f(x)) → PROPER(x)
    The graph contains the following edges 1 > 1

(19) YES

(20) Obligation:

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

MATCH(f(x), f(y)) → MATCH(x, y)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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

MATCH(f(x), f(y)) → MATCH(x, y)

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

(23) 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:

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

(24) YES

(25) Obligation:

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

CHECK(f(x)) → CHECK(x)

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(26) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(27) Obligation:

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

CHECK(f(x)) → CHECK(x)

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

(28) 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:

  • CHECK(f(x)) → CHECK(x)
    The graph contains the following edges 1 > 1

(29) YES

(30) Obligation:

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

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
TOP(active(c)) → TOP(mark(c))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
top(active(c)) → top(mark(c))
top(mark(x)) → top(check(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))
f(ok(x)) → ok(f(x))
start(ok(x)) → found(x)
f(found(x)) → found(f(x))
top(found(x)) → top(active(x))
active(f(x)) → f(active(x))
f(mark(x)) → mark(f(x))

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

(31) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(32) Obligation:

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

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))
TOP(active(c)) → TOP(mark(c))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
start(ok(x)) → found(x)
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


TOP(active(c)) → TOP(mark(c))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
mark(x1)  =  mark
check(x1)  =  check
found(x1)  =  x1
active(x1)  =  x1
c  =  c
f(x1)  =  f
start(x1)  =  x1
match(x1, x2)  =  x1
X  =  X
ok(x1)  =  x1
proper(x1)  =  proper

Recursive path order with status [RPO].
Quasi-Precedence:
TOP1 > [c, proper] > [mark, check, f]
X > [c, proper] > [mark, check, f]

Status:
TOP1: [1]
mark: multiset
check: multiset
c: multiset
f: multiset
X: multiset
proper: multiset


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
match(f(x), f(y)) → f(match(x, y))
start(ok(x)) → found(x)

(34) Obligation:

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

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
match(f(x), f(y)) → f(match(x, y))
start(ok(x)) → found(x)
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))

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

(35) 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 rules of the TRS R:

match(f(x), f(y)) → f(match(x, y))

Used ordering: Polynomial interpretation [POLO]:

POL(TOP(x1)) = x1   
POL(X) = 0   
POL(active(x1)) = x1   
POL(c) = 1   
POL(check(x1)) = 1 + x1   
POL(f(x1)) = 1 + x1   
POL(found(x1)) = x1   
POL(mark(x1)) = 1 + x1   
POL(match(x1, x2)) = x1 + x2   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   
POL(start(x1)) = x1   

(36) Obligation:

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

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))
start(ok(x)) → found(x)
match(X, x) → proper(x)
proper(c) → ok(c)
proper(f(x)) → f(proper(x))

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

(37) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(38) Obligation:

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

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))

The TRS R consists of the following rules:

active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))

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

(39) 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:

TOP(mark(x)) → TOP(check(x))
TOP(found(x)) → TOP(active(x))

Strictly oriented rules of the TRS R:

active(f(x)) → mark(x)
active(f(x)) → f(active(x))
f(ok(x)) → ok(f(x))
f(found(x)) → found(f(x))
f(mark(x)) → mark(f(x))
check(f(x)) → f(check(x))
check(x) → start(match(f(X), x))

Used ordering: Knuth-Bendix order [KBO] with precedence:
active1 > X > check1 > start1 > f1 > mark1 > TOP1 > match2 > found1 > ok1

and weight map:

X=1
active_1=3
f_1=5
mark_1=8
ok_1=1
found_1=4
check_1=7
start_1=1
TOP_1=1
match_2=0

The variable weight is 1

(40) Obligation:

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

(41) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(42) YES