(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(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:
fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
(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:
FAC(0) → 11
FAC(s(x)) → *1(s(x), fac(x))
FAC(s(x)) → FAC(x)
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
FLOOP(s(x), y) → *1(s(x), y)
*1(x, s(y)) → +1(*(x, y), x)
*1(x, s(y)) → *1(x, y)
+1(x, s(y)) → +1(x, y)
The TRS R consists of the following rules:
fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(x, s(y)) → +1(x, y)
The TRS R consists of the following rules:
fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
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:
+1(x, s(y)) → +1(x, y)
R is empty.
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
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].
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(x, s(y)) → +1(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:
- +1(x, s(y)) → +1(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(x, s(y)) → *1(x, y)
The TRS R consists of the following rules:
fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
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:
*1(x, s(y)) → *1(x, y)
R is empty.
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
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].
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(x, s(y)) → *1(x, y)
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:
- *1(x, s(y)) → *1(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(20) TRUE
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
The TRS R consists of the following rules:
fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
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:
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
The TRS R consists of the following rules:
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
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].
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
1
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
The TRS R consists of the following rules:
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
The set Q consists of the following terms:
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
We have to consider all minimal (P,Q,R)-chains.
(26) 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:
- FLOOP(s(x), y) → FLOOP(x, *(s(x), y))
The graph contains the following edges 1 > 1
(27) TRUE
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(x)
The TRS R consists of the following rules:
fac(0) → 1
fac(s(x)) → *(s(x), fac(x))
floop(0, y) → y
floop(s(x), y) → floop(x, *(s(x), y))
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
1 → s(0)
fac(0) → s(0)
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
We have to consider all minimal (P,Q,R)-chains.
(29) 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.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(x)
R is empty.
The set Q consists of the following terms:
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
We have to consider all minimal (P,Q,R)-chains.
(31) 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].
fac(0)
fac(s(x0))
floop(0, x0)
floop(s(x0), x1)
*(x0, 0)
*(x0, s(x1))
+(x0, 0)
+(x0, s(x1))
1
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) 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:
- FAC(s(x)) → FAC(x)
The graph contains the following edges 1 > 1
(34) TRUE