YES Termination w.r.t. Q proof of AProVE_07_thiemann38.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 1 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 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 NonInfProof (⇔, 0 ms)
        ↳27 AND
            ↳28 QDP
                ↳29 DependencyGraphProof (⇔, 0 ms)
                ↳30 TRUE
            ↳31 QDP
                ↳32 DependencyGraphProof (⇔, 0 ms)
                ↳33 TRUE

(0) Obligation:

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

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

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:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(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:

LENGTH(cons(x, l)) → LENGTH(l)
LT(s(x), s(y)) → LT(x, y)
REVERSE(l) → REV(0, l, nil, l)
REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
REV(x, l, accu, orig) → LT(x, length(orig))
REV(x, l, accu, orig) → LENGTH(orig)
IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)
IF(true, x, l, accu, orig) → TAIL(l)
IF(true, x, l, accu, orig) → HEAD(l)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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:

LT(s(x), s(y)) → LT(x, y)

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

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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].

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(11) Obligation:

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

LT(s(x), s(y)) → LT(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:

  • LT(s(x), s(y)) → LT(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:

LENGTH(cons(x, l)) → LENGTH(l)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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:

LENGTH(cons(x, l)) → LENGTH(l)

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

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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].

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(18) Obligation:

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

LENGTH(cons(x, l)) → LENGTH(l)

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

(19) 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:

  • LENGTH(cons(x, l)) → LENGTH(l)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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

(22) 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.

(23) Obligation:

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

REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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

(24) 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].

reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(25) Obligation:

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

REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))

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

(26) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig) the following chains were created:
  • We consider the chain IF(true, x4, x5, x6, x7) → REV(s(x4), tail(x5), cons(head(x5), x6), x7), REV(x8, x9, x10, x11) → IF(lt(x8, length(x11)), x8, x9, x10, x11) which results in the following constraint:
    (1)    (REV(s(x4), tail(x5), cons(head(x5), x6), x7)=REV(x8, x9, x10, x11) ⇒ REV(x8, x9, x10, x11)≥IF(lt(x8, length(x11)), x8, x9, x10, x11))


    We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (2)    (REV(s(x4), x9, x10, x7)≥IF(lt(s(x4), length(x7)), s(x4), x9, x10, x7))






For Pair IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig) the following chains were created:
  • We consider the chain REV(x12, x13, x14, x15) → IF(lt(x12, length(x15)), x12, x13, x14, x15), IF(true, x16, x17, x18, x19) → REV(s(x16), tail(x17), cons(head(x17), x18), x19) which results in the following constraint:
    (1)    (IF(lt(x12, length(x15)), x12, x13, x14, x15)=IF(true, x16, x17, x18, x19) ⇒ IF(true, x16, x17, x18, x19)≥REV(s(x16), tail(x17), cons(head(x17), x18), x19))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (length(x15)=x24∧lt(x12, x24)=true ⇒ IF(true, x12, x13, x14, x15)≥REV(s(x12), tail(x13), cons(head(x13), x14), x15))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x12, x24)=true which results in the following new constraints:
    (3)    (true=true∧length(x15)=s(x26) ⇒ IF(true, 0, x13, x14, x15)≥REV(s(0), tail(x13), cons(head(x13), x14), x15))

    (4)    (lt(x28, x27)=true∧length(x15)=s(x27)∧(∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29)) ⇒ IF(true, s(x28), x13, x14, x15)≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), x15))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (length(x15)=s(x26) ⇒ IF(true, 0, x13, x14, x15)≥REV(s(0), tail(x13), cons(head(x13), x14), x15))


    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x15)=s(x27) which results in the following new constraint:
    (6)    (s(length(x37))=s(x27)∧lt(x28, x27)=true∧(∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29))∧(∀x39,x40,x41,x42,x43,x44,x45:length(x37)=s(x39)∧lt(x40, x39)=true∧(∀x41,x42,x43:lt(x40, x39)=true∧length(x41)=x39 ⇒ IF(true, x40, x42, x43, x41)≥REV(s(x40), tail(x42), cons(head(x42), x43), x41)) ⇒ IF(true, s(x40), x44, x45, x37)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x37)) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))


    We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x15)=s(x26) which results in the following new constraint:
    (7)    (s(length(x32))=s(x26)∧(∀x34,x35,x36:length(x32)=s(x34) ⇒ IF(true, 0, x35, x36, x32)≥REV(s(0), tail(x35), cons(head(x35), x36), x32)) ⇒ IF(true, 0, x13, x14, cons(x33, x32))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x33, x32)))


    We simplified constraint (7) using rules (I), (II), (IV) which results in the following new constraint:
    (8)    (IF(true, 0, x13, x14, cons(x33, x32))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x33, x32)))


    We simplified constraint (6) using rules (I), (II) which results in the following new constraint:
    (9)    (length(x37)=x27∧lt(x28, x27)=true∧(∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29))∧(∀x39,x40,x41,x42,x43,x44,x45:length(x37)=s(x39)∧lt(x40, x39)=true∧(∀x41,x42,x43:lt(x40, x39)=true∧length(x41)=x39 ⇒ IF(true, x40, x42, x43, x41)≥REV(s(x40), tail(x42), cons(head(x42), x43), x41)) ⇒ IF(true, s(x40), x44, x45, x37)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x37)) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))


    We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29)) with σ = [x29 / x37, x30 / x13, x31 / x14] which results in the following new constraint:
    (10)    (IF(true, x28, x13, x14, x37)≥REV(s(x28), tail(x13), cons(head(x13), x14), x37)∧(∀x39,x40,x41,x42,x43,x44,x45:length(x37)=s(x39)∧lt(x40, x39)=true∧(∀x41,x42,x43:lt(x40, x39)=true∧length(x41)=x39 ⇒ IF(true, x40, x42, x43, x41)≥REV(s(x40), tail(x42), cons(head(x42), x43), x41)) ⇒ IF(true, s(x40), x44, x45, x37)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x37)) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))


    We simplified constraint (10) using rule (IV) which results in the following new constraint:
    (11)    (IF(true, x28, x13, x14, x37)≥REV(s(x28), tail(x13), cons(head(x13), x14), x37) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))






To summarize, we get the following constraints P≥ for the following pairs.
  • REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
    • (REV(s(x4), x9, x10, x7)≥IF(lt(s(x4), length(x7)), s(x4), x9, x10, x7))

  • IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)
    • (IF(true, 0, x13, x14, cons(x33, x32))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x33, x32)))
    • (IF(true, x28, x13, x14, x37)≥REV(s(x28), tail(x13), cons(head(x13), x14), x37) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))




The constraints for P> respective Pbound are constructed from P≥ where we just replace every occurence of "t ≥ s" in P≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 1   
POL(IF(x1, x2, x3, x4, x5)) = -1 - x2 + x5   
POL(REV(x1, x2, x3, x4)) = -x1 + x4   
POL(c) = -1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(false) = 1   
POL(head(x1)) = 0   
POL(length(x1)) = 1   
POL(lt(x1, x2)) = 1 + x2   
POL(nil) = 0   
POL(s(x1)) = 1 + x1   
POL(tail(x1)) = 0   
POL(true) = 1   
POL(undefined) = 1   

The following pairs are in P>:

REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
The following pairs are in Pbound:

IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)
There are no usable rules

(27) Complex Obligation (AND)

(28) Obligation:

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

IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))

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

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) TRUE

(31) Obligation:

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

REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))

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

(32) DependencyGraphProof (EQUIVALENT transformation)

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

(33) TRUE