(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)
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:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(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:
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → DIV(f(x, minus(s(y), s(0)), b), b)
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
F(x, s(y), b) → MINUS(s(y), s(0))
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(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 3 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(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:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
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].
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(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:
- MINUS(s(x), s(y)) → MINUS(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:
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
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:
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
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].
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(19) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(DIV(x1, x2)) = x1
POL(minus(x1, x2)) = x1
POL(s(x1)) = 1 + x1
The following usable rules [FROCOS05] were oriented:
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
minus(x, x) → 0
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) TRUE
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(24) 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.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(26) 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].
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(28) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
F(
x,
s(
y),
b) →
F(
x,
minus(
s(
y),
s(
0)),
b) at position [1] we obtained the following new rules [LPAR04]:
F(y0, s(0), y2) → F(y0, 0, y2)
F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y0, s(0), y2) → F(y0, 0, y2)
F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(30) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(32) 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.
(33) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(34) Rewriting (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
F(
y0,
s(
x0),
y2) →
F(
y0,
minus(
x0,
0),
y2) at position [1] we obtained the following new rules [LPAR04]:
F(y0, s(x0), y2) → F(y0, x0, y2)
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y0, s(x0), y2) → F(y0, x0, y2)
The TRS R consists of the following rules:
minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(36) 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.
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y0, s(x0), y2) → F(y0, x0, y2)
R is empty.
The set Q consists of the following terms:
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
(38) 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].
minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y0, s(x0), y2) → F(y0, x0, y2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(40) ForwardInstantiation (EQUIVALENT transformation)
By forward instantiating [JAR06] the rule
F(
y0,
s(
x0),
y2) →
F(
y0,
x0,
y2) we obtained the following new rules [LPAR04]:
F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(42) 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:
- F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
(43) TRUE