(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The TRS R 2 is
f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
The signature Sigma is {
f,
g}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
(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:
F(true, x, y, z) → G(gt(x, y), x, y, z)
F(true, x, y, z) → GT(x, y)
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
G(true, x, y, z) → GT(x, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
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 2 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
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:
GT(s(u), s(v)) → GT(u, v)
R is empty.
The set Q consists of the following terms:
f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
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].
f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GT(s(u), s(v)) → GT(u, v)
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:
- GT(s(u), s(v)) → GT(u, v)
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:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, x, y, z) → G(gt(x, y), x, y, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
The TRS R consists of the following rules:
f(true, x, y, z) → g(gt(x, y), x, y, z)
g(true, x, y, z) → f(gt(x, z), x, s(y), z)
g(true, x, y, z) → f(gt(x, z), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
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:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, x, y, z) → G(gt(x, y), x, y, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1, x2)
g(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
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].
f(true, x0, x1, x2)
g(true, x0, x1, x2)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, x, y, z) → G(gt(x, y), x, y, z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(19) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
F(
true,
x,
y,
z) →
G(
gt(
x,
y),
x,
y,
z) we obtained the following new rules [LPAR04]:
F(true, z0, s(z1), z2) → G(gt(z0, s(z1)), z0, s(z1), z2)
F(true, z0, z1, s(z2)) → G(gt(z0, z1), z0, z1, s(z2))
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
F(true, z0, s(z1), z2) → G(gt(z0, s(z1)), z0, s(z1), z2)
F(true, z0, z1, s(z2)) → G(gt(z0, z1), z0, z1, s(z2))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(21) 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
G(
true,
x,
y,
z) →
F(
gt(
x,
z),
x,
s(
y),
z) the following chains were created:
- We consider the chain F(true, x, y, z) → G(gt(x, y), x, y, z), G(true, x, y, z) → F(gt(x, z), x, s(y), z) which results in the following constraint:
(1) (G(gt(x3, x4), x3, x4, x5)=G(true, x6, x7, x8) ⇒ G(true, x6, x7, x8)≥F(gt(x6, x8), x6, s(x7), x8))
We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (gt(x3, x4)=true ⇒ G(true, x3, x4, x5)≥F(gt(x3, x5), x3, s(x4), x5))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x3, x4)=true which results in the following new constraints:
(3) (true=true ⇒ G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))
(4) (gt(x42, x41)=true∧(∀x43:gt(x42, x41)=true ⇒ G(true, x42, x41, x43)≥F(gt(x42, x43), x42, s(x41), x43)) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))
We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
(5) (G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))
We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x43:gt(x42, x41)=true ⇒ G(true, x42, x41, x43)≥F(gt(x42, x43), x42, s(x41), x43)) with σ = [x43 / x5] which results in the following new constraint:
(6) (G(true, x42, x41, x5)≥F(gt(x42, x5), x42, s(x41), x5) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))
For Pair
F(
true,
x,
y,
z) →
G(
gt(
x,
y),
x,
y,
z) the following chains were created:
- We consider the chain G(true, x, y, z) → F(gt(x, z), x, s(y), z), F(true, x, y, z) → G(gt(x, y), x, y, z) which results in the following constraint:
(7) (F(gt(x12, x14), x12, s(x13), x14)=F(true, x15, x16, x17) ⇒ F(true, x15, x16, x17)≥G(gt(x15, x16), x15, x16, x17))
We simplified constraint (7) using rules (I), (II), (III) which results in the following new constraint:
(8) (gt(x12, x14)=true ⇒ F(true, x12, s(x13), x14)≥G(gt(x12, s(x13)), x12, s(x13), x14))
We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on gt(x12, x14)=true which results in the following new constraints:
(9) (true=true ⇒ F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))
(10) (gt(x47, x46)=true∧(∀x48:gt(x47, x46)=true ⇒ F(true, x47, s(x48), x46)≥G(gt(x47, s(x48)), x47, s(x48), x46)) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))
We simplified constraint (9) using rules (I), (II) which results in the following new constraint:
(11) (F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))
We simplified constraint (10) using rule (VI) where we applied the induction hypothesis (∀x48:gt(x47, x46)=true ⇒ F(true, x47, s(x48), x46)≥G(gt(x47, s(x48)), x47, s(x48), x46)) with σ = [x48 / x13] which results in the following new constraint:
(12) (F(true, x47, s(x13), x46)≥G(gt(x47, s(x13)), x47, s(x13), x46) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))
- We consider the chain G(true, x, y, z) → F(gt(x, z), x, y, s(z)), F(true, x, y, z) → G(gt(x, y), x, y, z) which results in the following constraint:
(13) (F(gt(x21, x23), x21, x22, s(x23))=F(true, x24, x25, x26) ⇒ F(true, x24, x25, x26)≥G(gt(x24, x25), x24, x25, x26))
We simplified constraint (13) using rules (I), (II), (III) which results in the following new constraint:
(14) (gt(x21, x23)=true ⇒ F(true, x21, x22, s(x23))≥G(gt(x21, x22), x21, x22, s(x23)))
We simplified constraint (14) using rule (V) (with possible (I) afterwards) using induction on gt(x21, x23)=true which results in the following new constraints:
(15) (true=true ⇒ F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))
(16) (gt(x52, x51)=true∧(∀x53:gt(x52, x51)=true ⇒ F(true, x52, x53, s(x51))≥G(gt(x52, x53), x52, x53, s(x51))) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))
We simplified constraint (15) using rules (I), (II) which results in the following new constraint:
(17) (F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))
We simplified constraint (16) using rule (VI) where we applied the induction hypothesis (∀x53:gt(x52, x51)=true ⇒ F(true, x52, x53, s(x51))≥G(gt(x52, x53), x52, x53, s(x51))) with σ = [x53 / x22] which results in the following new constraint:
(18) (F(true, x52, x22, s(x51))≥G(gt(x52, x22), x52, x22, s(x51)) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))
For Pair
G(
true,
x,
y,
z) →
F(
gt(
x,
z),
x,
y,
s(
z)) the following chains were created:
- We consider the chain F(true, x, y, z) → G(gt(x, y), x, y, z), G(true, x, y, z) → F(gt(x, z), x, y, s(z)) which results in the following constraint:
(19) (G(gt(x30, x31), x30, x31, x32)=G(true, x33, x34, x35) ⇒ G(true, x33, x34, x35)≥F(gt(x33, x35), x33, x34, s(x35)))
We simplified constraint (19) using rules (I), (II), (III) which results in the following new constraint:
(20) (gt(x30, x31)=true ⇒ G(true, x30, x31, x32)≥F(gt(x30, x32), x30, x31, s(x32)))
We simplified constraint (20) using rule (V) (with possible (I) afterwards) using induction on gt(x30, x31)=true which results in the following new constraints:
(21) (true=true ⇒ G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))
(22) (gt(x57, x56)=true∧(∀x58:gt(x57, x56)=true ⇒ G(true, x57, x56, x58)≥F(gt(x57, x58), x57, x56, s(x58))) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))
We simplified constraint (21) using rules (I), (II) which results in the following new constraint:
(23) (G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))
We simplified constraint (22) using rule (VI) where we applied the induction hypothesis (∀x58:gt(x57, x56)=true ⇒ G(true, x57, x56, x58)≥F(gt(x57, x58), x57, x56, s(x58))) with σ = [x58 / x32] which results in the following new constraint:
(24) (G(true, x57, x56, x32)≥F(gt(x57, x32), x57, x56, s(x32)) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))
To summarize, we get the following constraints P
≥ for the following pairs.
- G(true, x, y, z) → F(gt(x, z), x, s(y), z)
- (G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))
- (G(true, x42, x41, x5)≥F(gt(x42, x5), x42, s(x41), x5) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))
- F(true, x, y, z) → G(gt(x, y), x, y, z)
- (F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))
- (F(true, x47, s(x13), x46)≥G(gt(x47, s(x13)), x47, s(x13), x46) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))
- (F(true, s(x50), x22, s(0))≥G(gt(s(x50), x22), s(x50), x22, s(0)))
- (F(true, x52, x22, s(x51))≥G(gt(x52, x22), x52, x22, s(x51)) ⇒ F(true, s(x52), x22, s(s(x51)))≥G(gt(s(x52), x22), s(x52), x22, s(s(x51))))
- G(true, x, y, z) → F(gt(x, z), x, y, s(z))
- (G(true, s(x55), 0, x32)≥F(gt(s(x55), x32), s(x55), 0, s(x32)))
- (G(true, x57, x56, x32)≥F(gt(x57, x32), x57, x56, s(x32)) ⇒ G(true, s(x57), s(x56), x32)≥F(gt(s(x57), x32), s(x57), s(x56), s(x32)))
The constraints for P
> respective P
bound 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 P
bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:
POL(0) = 0
POL(F(x1, x2, x3, x4)) = -1 - x1 + x2 - x4
POL(G(x1, x2, x3, x4)) = -1 - x1 + x2 - x4
POL(c) = -1
POL(false) = 1
POL(gt(x1, x2)) = 0
POL(s(x1)) = 1 + x1
POL(true) = 0
The following pairs are in P
>:
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
The following pairs are in P
bound:
F(true, x, y, z) → G(gt(x, y), x, y, z)
The following rules are usable:
true → gt(s(u), 0)
false → gt(0, v)
gt(u, v) → gt(s(u), s(v))
(22) Complex Obligation (AND)
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, x, y, z) → G(gt(x, y), x, y, z)
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(24) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
F(
true,
x,
y,
z) →
G(
gt(
x,
y),
x,
y,
z) we obtained the following new rules [LPAR04]:
F(true, z0, s(z1), z2) → G(gt(z0, s(z1)), z0, s(z1), z2)
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
F(true, z0, s(z1), z2) → G(gt(z0, s(z1)), z0, s(z1), z2)
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(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
G(
true,
x,
y,
z) →
F(
gt(
x,
z),
x,
s(
y),
z) the following chains were created:
- We consider the chain F(true, x, y, z) → G(gt(x, y), x, y, z), G(true, x, y, z) → F(gt(x, z), x, s(y), z) which results in the following constraint:
(1) (G(gt(x3, x4), x3, x4, x5)=G(true, x6, x7, x8) ⇒ G(true, x6, x7, x8)≥F(gt(x6, x8), x6, s(x7), x8))
We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (gt(x3, x4)=true ⇒ G(true, x3, x4, x5)≥F(gt(x3, x5), x3, s(x4), x5))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x3, x4)=true which results in the following new constraints:
(3) (true=true ⇒ G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))
(4) (gt(x42, x41)=true∧(∀x43:gt(x42, x41)=true ⇒ G(true, x42, x41, x43)≥F(gt(x42, x43), x42, s(x41), x43)) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))
We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
(5) (G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))
We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x43:gt(x42, x41)=true ⇒ G(true, x42, x41, x43)≥F(gt(x42, x43), x42, s(x41), x43)) with σ = [x43 / x5] which results in the following new constraint:
(6) (G(true, x42, x41, x5)≥F(gt(x42, x5), x42, s(x41), x5) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))
For Pair
F(
true,
x,
y,
z) →
G(
gt(
x,
y),
x,
y,
z) the following chains were created:
- We consider the chain G(true, x, y, z) → F(gt(x, z), x, s(y), z), F(true, x, y, z) → G(gt(x, y), x, y, z) which results in the following constraint:
(7) (F(gt(x12, x14), x12, s(x13), x14)=F(true, x15, x16, x17) ⇒ F(true, x15, x16, x17)≥G(gt(x15, x16), x15, x16, x17))
We simplified constraint (7) using rules (I), (II), (III) which results in the following new constraint:
(8) (gt(x12, x14)=true ⇒ F(true, x12, s(x13), x14)≥G(gt(x12, s(x13)), x12, s(x13), x14))
We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on gt(x12, x14)=true which results in the following new constraints:
(9) (true=true ⇒ F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))
(10) (gt(x47, x46)=true∧(∀x48:gt(x47, x46)=true ⇒ F(true, x47, s(x48), x46)≥G(gt(x47, s(x48)), x47, s(x48), x46)) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))
We simplified constraint (9) using rules (I), (II) which results in the following new constraint:
(11) (F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))
We simplified constraint (10) using rule (VI) where we applied the induction hypothesis (∀x48:gt(x47, x46)=true ⇒ F(true, x47, s(x48), x46)≥G(gt(x47, s(x48)), x47, s(x48), x46)) with σ = [x48 / x13] which results in the following new constraint:
(12) (F(true, x47, s(x13), x46)≥G(gt(x47, s(x13)), x47, s(x13), x46) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))
To summarize, we get the following constraints P
≥ for the following pairs.
- G(true, x, y, z) → F(gt(x, z), x, s(y), z)
- (G(true, s(x40), 0, x5)≥F(gt(s(x40), x5), s(x40), s(0), x5))
- (G(true, x42, x41, x5)≥F(gt(x42, x5), x42, s(x41), x5) ⇒ G(true, s(x42), s(x41), x5)≥F(gt(s(x42), x5), s(x42), s(s(x41)), x5))
- F(true, x, y, z) → G(gt(x, y), x, y, z)
- (F(true, s(x45), s(x13), 0)≥G(gt(s(x45), s(x13)), s(x45), s(x13), 0))
- (F(true, x47, s(x13), x46)≥G(gt(x47, s(x13)), x47, s(x13), x46) ⇒ F(true, s(x47), s(x13), s(x46))≥G(gt(s(x47), s(x13)), s(x47), s(x13), s(x46)))
The constraints for P
> respective P
bound 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 P
bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:
POL(0) = 0
POL(F(x1, x2, x3, x4)) = x1 + x2 - x3
POL(G(x1, x2, x3, x4)) = x2 - x3
POL(c) = -1
POL(false) = 0
POL(gt(x1, x2)) = 0
POL(s(x1)) = 1 + x1
POL(true) = 0
The following pairs are in P
>:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
The following pairs are in P
bound:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
The following rules are usable:
gt(s(u), 0) → true
gt(0, v) → false
gt(s(u), s(v)) → gt(u, v)
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y, z) → G(gt(x, y), x, y, z)
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(28) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(29) TRUE
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(true, x, y, z) → F(gt(x, z), x, s(y), z)
G(true, x, y, z) → F(gt(x, z), x, y, s(z))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(31) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(32) TRUE