YES Termination w.r.t. Q proof of AProVE_09_Inductive_zerolist.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 4 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 QDPOrderProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 Induction-Processor (⇒, 41 ms)
        ↳22 AND
            ↳23 QDP
                ↳24 DependencyGraphProof (⇔, 0 ms)
                ↳25 TRUE
            ↳26 QTRS
                ↳27 QTRSRRRProof (⇔, 0 ms)
                ↳28 QTRS
                ↳29 RisEmptyProof (⇔, 0 ms)
                ↳30 YES

(0) Obligation:

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

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

(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:

SUB(s(x), s(y)) → SUB(x, y)
ZERO(nil) → ZERO2(0, nil)
ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO(cons(x, xs)) → SUB(x, x)
ZERO2(0, cons(x, xs)) → SUB(x, x)
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), nil) → ZERO(nil)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

SUB(s(x), s(y)) → SUB(x, y)

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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:

SUB(s(x), s(y)) → SUB(x, y)

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

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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].

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

(11) Obligation:

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

SUB(s(x), s(y)) → SUB(x, 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:

  • SUB(s(x), s(y)) → SUB(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:

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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:

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))
zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

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].

zero(nil)
zero(cons(x0, x1))
zero2(0, nil)
zero2(0, cons(x0, x1))
zero2(s(x0), nil)
zero2(s(x0), cons(x1, x2))

(18) Obligation:

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

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(0, cons(x, xs)) → ZERO(xs)
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))

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

(19) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


ZERO2(0, cons(x, xs)) → ZERO(xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZERO(x1)  =  x1
cons(x1, x2)  =  cons(x2)
ZERO2(x1, x2)  =  x2

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

dummyConstant=1
cons_1=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(20) Obligation:

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

ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))
ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))

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

(21) Induction-Processor (SOUND transformation)


This DP could be deleted by the Induction-Processor:
ZERO(cons(x, xs)) → ZERO2(sub(x, x), cons(x, xs))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ZERO(x1)) = 1   
POL(ZERO2(x1, x2)) = x1   
POL(cons(x1, x2)) = 1   
POL(s(x1)) = 1   
POL(sub(x1, x2)) = 1   

At least one of these decreasing rules is always used after the deleted DP:
sub(0, 0) → 0
sub(0, s(x2)) → 0


The following formula is valid:
x:sort[a12].sub'(x , x )=true


The transformed set:
sub'(0, 0) → true
sub'(s(x''), s(y')) → sub'(x'', y')
sub'(s(x1), 0) → false
sub'(0, s(x2)) → true
sub(0, 0) → 0
sub(s(x''), s(y')) → sub(x'', y')
sub(s(x1), 0) → s(x1)
sub(0, s(x2)) → 0
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[a12](0, 0) → true
equal_sort[a12](0, s(v5)) → false
equal_sort[a12](s(v6), 0) → false
equal_sort[a12](s(v6), s(v7)) → equal_sort[a12](v6, v7)
equal_sort[a5](witness_sort[a5], witness_sort[a5]) → true
equal_sort[a19](cons(v8, v9), cons(v10, v11)) → and(equal_sort[a12](v8, v10), equal_sort[a5](v9, v11))
equal_sort[a19](cons(v8, v9), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(v12, v13)) → false
equal_sort[a19](witness_sort[a19], witness_sort[a19]) → true
equal_sort[a16](witness_sort[a16], witness_sort[a16]) → true


The proof given by the theorem prover:
The following output was given by the internal theorem prover:
proof of internal
# AProVE Commit ID: 3a20a6ef7432c3f292db1a8838479c42bf5e3b22 root 20240618 unpublished


Partial correctness of the following Program

   [x, v5, v6, v7, v8, v9, v10, v11, v12, v13, x'', y', x1, x2]
   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[a12](0, 0) -> true
   equal_sort[a12](0, s(v5)) -> false
   equal_sort[a12](s(v6), 0) -> false
   equal_sort[a12](s(v6), s(v7)) -> equal_sort[a12](v6, v7)
   equal_sort[a5](witness_sort[a5], witness_sort[a5]) -> true
   equal_sort[a19](cons(v8, v9), cons(v10, v11)) -> equal_sort[a12](v8, v10) and equal_sort[a5](v9, v11)
   equal_sort[a19](cons(v8, v9), witness_sort[a19]) -> false
   equal_sort[a19](witness_sort[a19], cons(v12, v13)) -> false
   equal_sort[a19](witness_sort[a19], witness_sort[a19]) -> true
   equal_sort[a16](witness_sort[a16], witness_sort[a16]) -> true
   sub'(0, 0) -> true
   sub'(s(x''), s(y')) -> sub'(x'', y')
   sub'(s(x1), 0) -> false
   sub'(0, s(x2)) -> true
   sub(0, 0) -> 0
   sub(s(x''), s(y')) -> sub(x'', y')
   sub(s(x1), 0) -> s(x1)
   sub(0, s(x2)) -> 0

using the following formula:
x:sort[a12].sub'(x, x)=true

could be successfully shown:
(0) Formula
(1) Induction by data structure [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) YES
    (6) Formula
        (7) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
        (8) YES


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

(0)
Obligation:
Formula:
x:sort[a12].sub'(x, x)=true

There are no hypotheses.




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

(1) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a12] generates the following cases:



1. Base Case:
Formula:
sub'(0, 0)=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a12].sub'(s(n), s(n))=true

Hypotheses:
n:sort[a12].sub'(n, n)=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
sub'(0, 0)=true

There are no hypotheses.




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

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

(5)
YES

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

(6)
Obligation:
Formula:
n:sort[a12].sub'(s(n), s(n))=true

Hypotheses:
n:sort[a12].sub'(n, n)=true




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

(7) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n:sort[a12].sub'(n, n)=true

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

(8)
YES

(22) Complex Obligation (AND)

(23) Obligation:

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

ZERO2(s(y), cons(x, xs)) → ZERO(cons(x, xs))

The TRS R consists of the following rules:

sub(0, 0) → 0
sub(s(x), s(y)) → sub(x, y)
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0

The set Q consists of the following terms:

sub(0, 0)
sub(s(x0), 0)
sub(0, s(x0))
sub(s(x0), s(x1))

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

(24) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(25) TRUE

(26) Obligation:

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

sub'(0, 0) → true
sub'(s(x''), s(y')) → sub'(x'', y')
sub'(s(x1), 0) → false
sub'(0, s(x2)) → true
sub(0, 0) → 0
sub(s(x''), s(y')) → sub(x'', y')
sub(s(x1), 0) → s(x1)
sub(0, s(x2)) → 0
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[a12](0, 0) → true
equal_sort[a12](0, s(v5)) → false
equal_sort[a12](s(v6), 0) → false
equal_sort[a12](s(v6), s(v7)) → equal_sort[a12](v6, v7)
equal_sort[a5](witness_sort[a5], witness_sort[a5]) → true
equal_sort[a19](cons(v8, v9), cons(v10, v11)) → and(equal_sort[a12](v8, v10), equal_sort[a5](v9, v11))
equal_sort[a19](cons(v8, v9), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(v12, v13)) → false
equal_sort[a19](witness_sort[a19], witness_sort[a19]) → true
equal_sort[a16](witness_sort[a16], witness_sort[a16]) → true

Q is empty.

(27) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 2   
POL(and(x1, x2)) = 2 + 2·x1 + x2   
POL(cons(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(equal_bool(x1, x2)) = 2 + x1 + x2   
POL(equal_sort[a12](x1, x2)) = 2 + x1 + x2   
POL(equal_sort[a16](x1, x2)) = 2 + 2·x1 + x2   
POL(equal_sort[a19](x1, x2)) = 2 + 2·x1 + x2   
POL(equal_sort[a5](x1, x2)) = 1 + x1 + 2·x2   
POL(false) = 2   
POL(isa_false(x1)) = 2 + 2·x1   
POL(isa_true(x1)) = 1 + 2·x1   
POL(not(x1)) = 2 + x1   
POL(or(x1, x2)) = 2 + 2·x1 + x2   
POL(s(x1)) = 1 + x1   
POL(sub(x1, x2)) = 2 + 2·x1 + x2   
POL(sub'(x1, x2)) = 2 + 2·x1 + x2   
POL(true) = 1   
POL(witness_sort[a16]) = 2   
POL(witness_sort[a19]) = 2   
POL(witness_sort[a5]) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

sub'(0, 0) → true
sub'(s(x''), s(y')) → sub'(x'', y')
sub'(s(x1), 0) → false
sub'(0, s(x2)) → true
sub(0, 0) → 0
sub(s(x''), s(y')) → sub(x'', y')
sub(s(x1), 0) → s(x1)
sub(0, s(x2)) → 0
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[a12](0, 0) → true
equal_sort[a12](0, s(v5)) → false
equal_sort[a12](s(v6), 0) → false
equal_sort[a12](s(v6), s(v7)) → equal_sort[a12](v6, v7)
equal_sort[a5](witness_sort[a5], witness_sort[a5]) → true
equal_sort[a19](cons(v8, v9), cons(v10, v11)) → and(equal_sort[a12](v8, v10), equal_sort[a5](v9, v11))
equal_sort[a19](cons(v8, v9), witness_sort[a19]) → false
equal_sort[a19](witness_sort[a19], cons(v12, v13)) → false
equal_sort[a19](witness_sort[a19], witness_sort[a19]) → true
equal_sort[a16](witness_sort[a16], witness_sort[a16]) → true


(28) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(29) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(30) YES