Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
IF(true, x, y, z) → P(z)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(x, y, z) → PLUS(z, s(0))
QUOT(x, y, z) → ZERO(x)
PLUS(s(x), y) → PLUS(x, s(y))
IF(false, x, s(y), z) → MINUS(x, s(y))
DIV(x, y) → QUOT(x, y, 0)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
IF(true, x, y, z) → P(z)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(x, y, z) → PLUS(z, s(0))
QUOT(x, y, z) → ZERO(x)
PLUS(s(x), y) → PLUS(x, s(y))
IF(false, x, s(y), z) → MINUS(x, s(y))
DIV(x, y) → QUOT(x, y, 0)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(x, s(y))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(x, s(y))
R is empty.
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(x, s(y))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- PLUS(s(x), y) → PLUS(x, s(y))
The graph contains the following edges 1 > 1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
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(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
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(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
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.
By using the subterm criterion [20] together with the size-change analysis [32] 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
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0))) at position [0] we obtained the following new rules:
QUOT(0, y1, y2) → IF(true, 0, y1, plus(y2, s(0)))
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(0, y1, y2) → IF(true, 0, y1, plus(y2, s(0)))
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
zero(s(x0))
zero(0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z) at position [0] we obtained the following new rules:
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
IF(false, 0, s(y1), y2) → QUOT(0, s(y1), y2)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
IF(false, 0, s(y1), y2) → QUOT(0, s(y1), y2)
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0))) we obtained the following new rules:
QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
The remaining pairs can at least be oriented weakly.
QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( plus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( minus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( QUOT(x1, ..., x3) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| M( IF(x1, ..., x4) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 | + | | · | x4 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
minus(s(x), s(y)) → minus(x, y)
minus(0, y) → 0
minus(x, 0) → x
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))
The TRS R consists of the following rules:
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
The set Q consists of the following terms:
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.