YES Termination w.r.t. Q proof of AotoYamada_05_016.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 (⇔, 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 ATransformationProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QReductionProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QReductionProof (⇔, 8 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES

(0) Obligation:

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

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

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:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

(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:

APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)
APP(app(neq, app(s, x)), app(s, y)) → APP(neq, x)
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(filter, f), app(app(cons, y), ys)) → APP(app(filtersub, app(f, y)), f)
APP(app(filter, f), app(app(cons, y), ys)) → APP(filtersub, app(f, y))
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(cons, y), app(app(filter, f), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(filter, f)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(filter, f)
NONZERO → APP(filter, app(neq, 0))
NONZERO → APP(neq, 0)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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:

APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)

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

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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].

nonzero

(11) Obligation:

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

APP(app(neq, app(s, x)), app(s, y)) → APP(app(neq, x), y)

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

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))

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

(12) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(13) Obligation:

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

neq1(s(x), s(y)) → neq1(x, y)

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

neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
filter(x0, nil)
filter(x0, cons(x1, x2))
filtersub(true, x0, cons(x1, x2))
filtersub(false, x0, cons(x1, x2))

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

(14) 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].

neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
filter(x0, nil)
filter(x0, cons(x1, x2))
filtersub(true, x0, cons(x1, x2))
filtersub(false, x0, cons(x1, x2))

(15) Obligation:

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

neq1(s(x), s(y)) → neq1(x, y)

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

(16) 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:

  • neq1(s(x), s(y)) → neq1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(17) YES

(18) Obligation:

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
nonzero → app(filter, app(neq, 0))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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

(19) 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.

(20) Obligation:

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))
nonzero

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

(21) 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].

nonzero

(22) Obligation:

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

APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)

The TRS R consists of the following rules:

app(app(neq, 0), 0) → false
app(app(neq, 0), app(s, y)) → true
app(app(neq, app(s, x)), 0) → true
app(app(neq, app(s, x)), app(s, y)) → app(app(neq, x), y)
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, y), ys)) → app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) → app(app(filter, f), ys)
app(app(app(filtersub, true), f), app(app(cons, y), ys)) → app(app(cons, y), app(app(filter, f), ys))

The set Q consists of the following terms:

app(app(neq, 0), 0)
app(app(neq, 0), app(s, x0))
app(app(neq, app(s, x0)), 0)
app(app(neq, app(s, x0)), app(s, x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(filtersub, true), x0), app(app(cons, x1), x2))
app(app(app(filtersub, false), x0), app(app(cons, x1), x2))

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:

  • APP(app(filter, f), app(app(cons, y), ys)) → APP(f, y)
    The graph contains the following edges 1 > 1, 2 > 2

  • APP(app(filter, f), app(app(cons, y), ys)) → APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
    The graph contains the following edges 2 >= 2

  • APP(app(app(filtersub, true), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
    The graph contains the following edges 2 > 2

  • APP(app(app(filtersub, false), f), app(app(cons, y), ys)) → APP(app(filter, f), ys)
    The graph contains the following edges 2 > 2

(24) YES