YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QReductionProof [EQUIVALENT, 1 ms]
(8) QDP
(9) TransformationProof [EQUIVALENT, 0 ms]
(10) QDP
(11) Induction-Processor [SOUND, 32 ms]
(12) AND
    (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QTRS
        (17) QTRSRRRProof [EQUIVALENT, 0 ms]
        (18) QTRS
        (19) RisEmptyProof [EQUIVALENT, 0 ms]
        (20) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(0, 1, X) -> f(g(X, X), X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   f(0, 1, x0)
   g(x0, x1)


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(1) DependencyPairsProof (EQUIVALENT)
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(0, 1, X) -> F(g(X, X), X, X)
   F(0, 1, X) -> G(X, X)

The TRS R consists of the following rules:

   f(0, 1, X) -> f(g(X, X), X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   f(0, 1, x0)
   g(x0, x1)

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, 1, X) -> F(g(X, X), X, X)

The TRS R consists of the following rules:

   f(0, 1, X) -> f(g(X, X), X, X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   f(0, 1, x0)
   g(x0, x1)

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

(5) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, 1, X) -> F(g(X, X), X, X)

The TRS R consists of the following rules:

   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   f(0, 1, x0)
   g(x0, x1)

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

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

   f(0, 1, x0)


----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, 1, X) -> F(g(X, X), X, X)

The TRS R consists of the following rules:

   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   g(x0, x1)

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

(9) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(0, 1, X) -> F(g(X, X), X, X) we obtained the following new rules [LPAR04]:

   (F(0, 1, 1) -> F(g(1, 1), 1, 1),F(0, 1, 1) -> F(g(1, 1), 1, 1))


----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, 1, 1) -> F(g(1, 1), 1, 1)

The TRS R consists of the following rules:

   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   g(x0, x1)

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

(11) Induction-Processor (SOUND)

This DP could be deleted by the Induction-Processor:
F(0, 1, 1) -> F(g(1, 1), 1, 1)


This order was computed:
Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(1) = 0
   POL(F(x_1, x_2, x_3)) = x_1
   POL(g(x_1, x_2)) = 1 + x_1 + x_2

At least one of these decreasing rules is always used after the deleted DP:
g(X, Y) -> X
g(X', Y') -> Y'


The following formula is valid:
z0:sort[a0].(~(z0=0)->g'(z0, z0)=true)


The transformed set:
g'(X, Y) -> true
g(X, Y) -> X
g(X', Y') -> Y'
equal_bool(true, false) -> false
equal_bool(false, true) -> false
equal_bool(true, true) -> true
equal_bool(false, false) -> true
and(true, x) -> x
and(false, x) -> false
or(true, x) -> true
or(false, x) -> x
not(false) -> true
not(true) -> false
isa_true(true) -> true
isa_true(false) -> false
isa_false(true) -> false
isa_false(false) -> true
equal_sort[a0](0, 0) -> true
equal_sort[a0](0, 1) -> false
equal_sort[a0](1, 0) -> false
equal_sort[a0](1, 1) -> true
equal_sort[a10](witness_sort[a10], witness_sort[a10]) -> true


The proof given by the theorem prover:
The following output was given by the internal theorem prover:proof of internal
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Partial correctness of the following Program

   [x, X, Y, X', Y']
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, 1) -> false
   equal_sort[a0](1, 0) -> false
   equal_sort[a0](1, 1) -> true
   equal_sort[a10](witness_sort[a10], witness_sort[a10]) -> true
   g'(X, Y) -> true
   g(X, Y) -> X
   g(X', Y') -> Y'

using the following formula:
z0:sort[a0].(~(z0=0)->g'(z0, z0)=true)

could be successfully shown:
(0) Formula
(1) Symbolic evaluation [EQUIVALENT, 0 ms]
(2) YES


----------------------------------------

(0)
Obligation:
Formula:
z0:sort[a0].(~(z0=0)->g'(z0, z0)=true)

There are no hypotheses.




----------------------------------------

(1) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(2)
YES

----------------------------------------

(12)
Complex Obligation (AND)

----------------------------------------

(13)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   g(x0, x1)

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

(14) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(15)
YES

----------------------------------------

(16)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g'(X, Y) -> true
   g(X, Y) -> X
   g(X', Y') -> Y'
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   and(true, x) -> x
   and(false, x) -> false
   or(true, x) -> true
   or(false, x) -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, 1) -> false
   equal_sort[a0](1, 0) -> false
   equal_sort[a0](1, 1) -> true
   equal_sort[a10](witness_sort[a10], witness_sort[a10]) -> true

Q is empty.

----------------------------------------

(17) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:isa_false_1 > witness_sort[a10] > equal_sort[a10]_2 > 1 > 0 > equal_sort[a0]_2 > isa_true_1 > equal_bool_2 > not_1 > or_2 > and_2 > g'_2 > true > false > g_2

and weight map:

   true=2
   false=2
   0=1
   1=1
   witness_sort[a10]=1
   not_1=1
   isa_true_1=1
   isa_false_1=0
   g'_2=0
   g_2=0
   equal_bool_2=0
   and_2=0
   or_2=0
   equal_sort[a0]_2=0
   equal_sort[a10]_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   g'(X, Y) -> true
   g(X, Y) -> X
   g(X', Y') -> Y'
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   and(true, x) -> x
   and(false, x) -> false
   or(true, x) -> true
   or(false, x) -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, 1) -> false
   equal_sort[a0](1, 0) -> false
   equal_sort[a0](1, 1) -> true
   equal_sort[a10](witness_sort[a10], witness_sort[a10]) -> true




----------------------------------------

(18)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

----------------------------------------

(19) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
----------------------------------------

(20)
YES