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:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
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:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(x, y, z) → INC(z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
DIVISION(x, y) → DIV(x, y, 0)
INC(s(x)) → INC(x)
IF(false, x, s(y), z) → MINUS(x, s(y))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → LT(x, y)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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:
LT(s(x), s(y)) → LT(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(x, y, z) → INC(z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
DIVISION(x, y) → DIV(x, y, 0)
INC(s(x)) → INC(x)
IF(false, x, s(y), z) → MINUS(x, s(y))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → LT(x, y)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 4 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INC(s(x)) → INC(x)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INC(s(x)) → INC(x)
R is empty.
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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.
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INC(s(x)) → INC(x)
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:
- INC(s(x)) → INC(x)
The graph contains the following edges 1 > 1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(x, y)
R is empty.
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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.
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(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:
- LT(s(x), s(y)) → LT(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
↳ 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:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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
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:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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.
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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.
division(x0, x1)
div(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
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule DIV(x, y, z) → IF(lt(x, y), x, y, inc(z)) at position [0] we obtained the following new rules:
DIV(0, s(x0), y2) → IF(true, 0, s(x0), inc(y2))
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
DIV(x0, 0, y2) → IF(false, x0, 0, inc(y2))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
DIV(x0, 0, y2) → IF(false, x0, 0, inc(y2))
DIV(0, s(x0), y2) → IF(true, 0, s(x0), inc(y2))
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z) at position [0] we obtained the following new rules:
IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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.
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
The remaining pairs can at least be oriented weakly.
IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( minus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( lt(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( DIV(x1, ..., x3) ) = | 1 | + | | · | 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(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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.