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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → 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:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
IF3(true, x, y) → MINUS(x, y)
MOD(x, y) → ID(y)
ID(s(x)) → ID(x)
MOD(x, y) → ZERO(x)
LE(s(x), s(y)) → LE(x, y)
MOD(x, y) → LE(y, x)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → ID(x)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
MOD(x, y) → ZERO(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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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:
IF3(true, x, y) → MINUS(x, y)
MOD(x, y) → ID(y)
ID(s(x)) → ID(x)
MOD(x, y) → ZERO(x)
LE(s(x), s(y)) → LE(x, y)
MOD(x, y) → LE(y, x)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → ID(x)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
MOD(x, y) → ZERO(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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 6 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:
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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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:
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))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
↳ 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:
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ID(s(x)) → ID(x)
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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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:
ID(s(x)) → ID(x)
R is empty.
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
↳ 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:
ID(s(x)) → ID(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:
- ID(s(x)) → ID(x)
The graph contains the following edges 1 > 1
↳ 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:
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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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:
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))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
↳ 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:
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
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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:
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
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, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)
↳ 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:
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y)) at position [0] we obtained the following new rules:
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1))
MOD(0, y1) → IF_MOD(true, zero(y1), le(y1, 0), id(0), id(y1))
↳ 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:
IF3(true, x, y) → MOD(minus(x, y), s(y))
MOD(0, y1) → IF_MOD(true, zero(y1), le(y1, 0), id(0), id(y1))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
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
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), id(s(x0)), id(y1)) at position [3] we obtained the following new rules:
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF3(true, x, y) → MOD(minus(x, y), s(y)) at position [0] we obtained the following new rules:
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF3(true, x0, 0) → MOD(x0, s(0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF3(true, x0, 0) → MOD(x0, s(0))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y) we obtained the following new rules:
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF2(false, b2, x, y) → IF3(b2, x, y) we obtained the following new rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1))
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule MOD(s(x0), y1) → IF_MOD(false, zero(y1), le(y1, s(x0)), s(id(x0)), id(y1)) we obtained the following new rules:
MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0)))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(0) → true
zero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
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
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0)))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(s(x)) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(0)) → IF_MOD(false, zero(s(0)), le(s(0), s(x0)), s(id(x0)), id(s(0))) at position [1] we obtained the following new rules:
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(s(x)) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, zero(s(s(z1))), le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1)))) at position [1] we obtained the following new rules:
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
zero(s(x)) → false
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
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
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
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(0)
zero(s(x0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, le(s(0), s(x0)), s(id(x0)), id(s(0))) at position [2] we obtained the following new rules:
MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0)))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(s(z1)), s(x0)), s(id(x0)), id(s(s(z1)))) at position [2] we obtained the following new rules:
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0)))
IF3(true, x0, 0) → MOD(x0, s(0))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1))))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, le(0, x0), s(id(x0)), id(s(0))) at position [2] we obtained the following new rules:
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0)))
IF3(true, x0, 0) → MOD(x0, s(0))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1))))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), id(s(s(z1)))) at position [4] we obtained the following new rules:
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1))))
IF3(true, x0, 0) → MOD(x0, s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), id(s(0))) at position [4] we obtained the following new rules:
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1))))
IF3(true, x0, 0) → MOD(x0, s(0))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0)))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(id(s(z1)))) at position [4,0] we obtained the following new rules:
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF3(true, x0, 0) → MOD(x0, s(0))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0)))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(id(0))) at position [4,0] we obtained the following new rules:
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF3(true, x0, 0) → MOD(x0, s(0))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF3(true, x0, 0) → MOD(x0, s(0)) we obtained the following new rules:
IF3(true, s(z1), 0) → MOD(s(z1), s(0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF3(true, s(z1), 0) → MOD(s(z1), s(0))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_MOD(false, y_0, y_1, s(y_2), y_3) → IF2(y_0, y_1, s(y_2), y_3) we obtained the following new rules:
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF_MOD(false, false, true, s(y_0), s(0)) → IF2(false, true, s(y_0), s(0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3)
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF_MOD(false, false, true, s(y_0), s(0)) → IF2(false, true, s(y_0), s(0))
IF3(true, s(z1), 0) → MOD(s(z1), s(0))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF2(false, z1, s(z2), z3) → IF3(z1, s(z2), z3) we obtained the following new rules:
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF2(false, true, s(z0), s(0)) → IF3(true, s(z0), s(0))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF_MOD(false, false, true, s(y_0), s(0)) → IF2(false, true, s(y_0), s(0))
IF3(true, s(z1), 0) → MOD(s(z1), s(0))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
IF2(false, true, s(z0), s(0)) → IF3(true, s(z0), s(0))
MOD(s(x0), s(0)) → IF_MOD(false, false, true, s(id(x0)), s(0))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
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 4 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1)))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF3(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(s(x1))) we obtained the following new rules:
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1))))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule MOD(s(x0), s(s(z1))) → IF_MOD(false, false, le(s(z1), x0), s(id(x0)), s(s(id(z1)))) we obtained the following new rules:
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(id(s(z1)))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(id(s(z1)))))
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(id(s(z1))))) at position [4,0,0] we obtained the following new rules:
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2)))
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF2(false, z0, s(z1), s(s(z2))) → IF3(z0, s(z1), s(s(z2))) we obtained the following new rules:
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF_MOD(false, false, y_0, s(y_1), s(s(y_2))) → IF2(false, y_0, s(y_1), s(s(y_2))) we obtained the following new rules:
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1)))))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(s(x0), s(s(s(z1)))) → IF_MOD(false, false, le(s(s(z1)), x0), s(id(x0)), s(s(s(id(z1))))) at position [2] we obtained the following new rules:
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1)))))
MOD(s(0), s(s(s(y1)))) → IF_MOD(false, false, false, s(id(0)), s(s(s(id(y1)))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1)))))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
MOD(s(0), s(s(s(y1)))) → IF_MOD(false, false, false, s(id(0)), s(s(s(id(y1)))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
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
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1)))))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(id(s(x1))), s(s(s(id(y1))))) at position [3,0] we obtained the following new rules:
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(s(id(x1))), s(s(s(id(y1)))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(s(id(x1))), s(s(s(id(y1)))))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
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.
MOD(s(s(x1)), s(s(s(y1)))) → IF_MOD(false, false, le(s(y1), x1), s(s(id(x1))), s(s(s(id(y1)))))
The remaining pairs can at least be oriented weakly.
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( minus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( le(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( IF3(x1, ..., x3) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| M( MOD(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
| M( IF2(x1, ..., x4) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 | + | | · | x4 |
| M( IF_MOD(x1, ..., x5) ) = | 0 | + | | · | 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:
id(0) → 0
id(s(x)) → s(id(x))
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
↳ Rewriting
↳ QDP
↳ Narrowing
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Rewriting
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Rewriting
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IF_MOD(false, false, true, s(x1), s(s(x2))) → IF2(false, true, s(x1), s(s(x2)))
IF2(false, true, s(x1), s(s(x2))) → IF3(true, s(x1), s(s(x2)))
IF3(true, s(z1), s(s(z2))) → MOD(minus(z1, s(z2)), s(s(s(z2))))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(s(x), s(y)) → le(x, y)
id(0) → 0
id(s(x)) → s(id(x))
le(0, y) → true
le(s(x), 0) → false
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
id(0)
id(s(x0))
minus(x0, 0)
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 3 less nodes.