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:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
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:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
MOD(x, y) → MINUS(x, y)
MOD(x, y) → LE(y, x)
MINUS(s(x), s(y)) → MINUS(x, y)
MOD(x, y) → ISZERO(y)
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
MOD(x, y) → MINUS(x, y)
MOD(x, y) → LE(y, x)
MINUS(s(x), s(y)) → MINUS(x, y)
MOD(x, y) → ISZERO(y)
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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 3 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:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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.
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
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
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
LE(s(x), s(y)) → LE(x, y)
R is empty.
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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.
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
LE(s(x), s(y)) → LE(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:
- LE(s(x), s(y)) → LE(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:
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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.
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y)) at position [0] we obtained the following new rules:
MOD(y0, 0) → IF_MOD(true, le(0, y0), y0, 0, minus(y0, 0))
MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
↳ 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:
MOD(y0, 0) → IF_MOD(true, le(0, y0), y0, 0, minus(y0, 0))
MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(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 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:
MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(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.
isZero(0)
isZero(s(x0))
↳ 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:
MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0))) at position [1] we obtained the following new rules:
MOD(0, s(x0)) → IF_MOD(false, false, 0, s(x0), minus(0, s(x0)))
MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0)))
↳ 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:
MOD(0, s(x0)) → IF_MOD(false, false, 0, s(x0), minus(0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)
MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0)))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(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
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, true, x, y, z) → MOD(z, y)
MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0)))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0))) at position [4] we obtained the following new rules:
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 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
↳ Narrowing
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)
IF_MOD(false, true, x, y, z) → MOD(z, y)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_MOD(false, true, x, y, z) → MOD(z, y) we obtained the following new rules:
IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1))
IF_MOD(false, true, s(z0), s(z0), 0) → MOD(0, s(z0))
↳ 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
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)
IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1))
IF_MOD(false, true, s(z0), s(z0), 0) → MOD(0, s(z0))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(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
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)
IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1)) we obtained the following new rules:
IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))
↳ 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
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(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
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(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_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))
The remaining pairs can at least be oriented weakly.
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( minus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( le(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( MOD(x1, x2) ) = | 1 | + | | · | x1 | + | | · | x2 |
| M( IF_MOD(x1, ..., x5) ) = | 1 | + | | · | x1 | + | | · | x2 | + | | · | x3 | + | | · | x4 | + | | · | x5 |
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(0, x) → 0
minus(x, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
↳ 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
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
le(0, y) → true
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(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.