(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))
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, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(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:
MINUS(s(X), s(Y)) → P(minus(X, Y))
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)
The TRS R consists of the following rules:
minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(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:
MINUS(s(X), s(Y)) → MINUS(X, Y)
The TRS R consists of the following rules:
minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(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:
MINUS(s(X), s(Y)) → MINUS(X, Y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(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].
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))
(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, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(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:
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
The TRS R consists of the following rules:
minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(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].
div(0, s(x0))
div(s(x0), s(x1))
(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, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
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(p(x1)) = x1
POL(s(x1)) = 1 + x1
The following usable rules [FROCOS05] were oriented:
minus(X, 0) → X
p(s(X)) → X
minus(s(X), s(Y)) → p(minus(X, Y))
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
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