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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 ATransformationProof (⇔)
    ↳11 QDP
    ↳12 QReductionProof (⇔)
    ↳13 QDP
    ↳14 QDPSizeChangeProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 UsableRulesProof (⇔)
    ↳18 QDP
    ↳19 ATransformationProof (⇔)
    ↳20 QDP
    ↳21 QReductionProof (⇔)
    ↳22 QDP
    ↳23 QDPSizeChangeProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 UsableRulesProof (⇔)
    ↳27 QDP
    ↳28 ATransformationProof (⇔)
    ↳29 QDP
    ↳30 QReductionProof (⇔)
    ↳31 QDP
    ↳32 QDPSizeChangeProof (⇔)
    ↳33 TRUE
  ↳34 QDP
    ↳35 UsableRulesProof (⇔)
    ↳36 QDP
    ↳37 ATransformationProof (⇔)
    ↳38 QDP
    ↳39 QReductionProof (⇔)
    ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 PisEmptyProof (⇔)
    ↳44 TRUE
  ↳45 QDP
    ↳46 UsableRulesProof (⇔)
    ↳47 QDP
    ↳48 QDPSizeChangeProof (⇔)
    ↳49 TRUE

(0) Obligation:

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

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), 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:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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:

APP(app(sub, app(s, x)), app(s, y)) → APP(app(sub, x), y)
APP(app(sub, app(s, x)), app(s, y)) → APP(sub, x)
APP(app(gtr, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(gtr, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
APP(app(d, app(s, x)), app(s, y)) → APP(app(if, app(app(gtr, x), y)), false)
APP(app(d, app(s, x)), app(s, y)) → APP(if, app(app(gtr, x), y))
APP(app(d, app(s, x)), app(s, y)) → APP(app(gtr, x), y)
APP(app(d, app(s, x)), app(s, y)) → APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))
APP(app(d, app(s, x)), app(s, y)) → APP(app(sub, y), x)
APP(app(d, app(s, x)), app(s, y)) → APP(sub, y)
APP(len, app(app(cons, x), xs)) → APP(s, app(len, xs))
APP(len, app(app(cons, x), xs)) → APP(len, xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs)))
APP(app(filter, p), app(app(cons, x), xs)) → APP(if, app(p, x))
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, p), xs))
APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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 5 SCCs with 14 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(len, app(app(cons, x), xs)) → APP(len, xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(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:

APP(len, app(app(cons, x), xs)) → APP(len, xs)

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(10) ATransformationProof (EQUIVALENT transformation)

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

(11) Obligation:

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

len1(cons(x, xs)) → len1(xs)

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

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

(12) 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].

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

(13) Obligation:

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

len1(cons(x, xs)) → len1(xs)

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

(14) 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:

  • len1(cons(x, xs)) → len1(xs)
    The graph contains the following edges 1 > 1

(15) TRUE

(16) Obligation:

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

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(17) 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.

(18) Obligation:

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

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

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(19) ATransformationProof (EQUIVALENT transformation)

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

(20) Obligation:

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

gtr1(s(x), s(y)) → gtr1(x, y)

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

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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].

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

(22) Obligation:

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

gtr1(s(x), s(y)) → gtr1(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:

  • gtr1(s(x), s(y)) → gtr1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(24) TRUE

(25) Obligation:

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

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

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(26) 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.

(27) Obligation:

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

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

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(28) ATransformationProof (EQUIVALENT transformation)

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

(29) Obligation:

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

sub1(s(x), s(y)) → sub1(x, y)

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

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

(30) 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].

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

(31) Obligation:

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

sub1(s(x), s(y)) → sub1(x, y)

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

(32) 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:

  • sub1(s(x), s(y)) → sub1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(33) TRUE

(34) Obligation:

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

APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(35) 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.

(36) Obligation:

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

APP(app(d, app(s, x)), app(s, y)) → APP(app(d, app(s, x)), app(app(sub, y), x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(37) ATransformationProof (EQUIVALENT transformation)

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

(38) Obligation:

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

d1(s(x), s(y)) → d1(s(x), sub(y, x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

if(true, x0, x1)
if(false, x0, x1)
sub(x0, 0)
sub(s(x0), s(x1))
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

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

(39) 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].

if(true, x0, x1)
if(false, x0, x1)
gtr(0, x0)
gtr(s(x0), 0)
gtr(s(x0), s(x1))
d(x0, 0)
d(s(x0), s(x1))
len(nil)
len(cons(x0, x1))
filter(x0, nil)
filter(x0, cons(x1, x2))

(40) Obligation:

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

d1(s(x), s(y)) → d1(s(x), sub(y, x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


d1(s(x), s(y)) → d1(s(x), sub(y, x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(d1(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   
POL(sub(x1, x2)) = x1   

The following usable rules [FROCOS05] were oriented:

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

(42) Obligation:

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

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

The set Q consists of the following terms:

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

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

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) TRUE

(45) Obligation:

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

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(sub, x), 0) → x
app(app(sub, app(s, x)), app(s, y)) → app(app(sub, x), y)
app(app(gtr, 0), y) → false
app(app(gtr, app(s, x)), 0) → true
app(app(gtr, app(s, x)), app(s, y)) → app(app(gtr, x), y)
app(app(d, x), 0) → true
app(app(d, app(s, x)), app(s, y)) → app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) → 0
app(len, app(app(cons, x), xs)) → app(s, app(len, xs))
app(app(filter, p), nil) → nil
app(app(filter, p), app(app(cons, x), xs)) → app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(46) 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.

(47) Obligation:

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

APP(app(filter, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)

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

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(sub, x0), 0)
app(app(sub, app(s, x0)), app(s, x1))
app(app(gtr, 0), x0)
app(app(gtr, app(s, x0)), 0)
app(app(gtr, app(s, x0)), app(s, x1))
app(app(d, x0), 0)
app(app(d, app(s, x0)), app(s, x1))
app(len, nil)
app(len, app(app(cons, x0), x1))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

(48) 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, p), app(app(cons, x), xs)) → APP(app(filter, p), xs)
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(filter, p), app(app(cons, x), xs)) → APP(p, x)
    The graph contains the following edges 1 > 1, 2 > 2

(49) TRUE