(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
B(x) → W(V(x))
M(x) → x
M(V(a(x))) → V(Xa(x))
M(V(b(x))) → V(Xb(x))
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xa(E(x)) → a(E(x))
Xb(E(x)) → b(E(x))
W(V(x)) → R(L(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(a(x)) → D(b(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
R(D(x)) → B(x)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(B(x1)) = 2 + x1
POL(D(x1)) = 3 + x1
POL(E(x1)) = x1
POL(L(x1)) = x1
POL(M(x1)) = 2 + x1
POL(R(x1)) = x1
POL(V(x1)) = 1 + x1
POL(W(x1)) = x1
POL(Xa(x1)) = 5 + x1
POL(Xb(x1)) = 1 + x1
POL(Ya(x1)) = 4 + x1
POL(Yb(x1)) = x1
POL(a(x1)) = 4 + x1
POL(b(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
B(x) → W(V(x))
M(x) → x
M(V(a(x))) → V(Xa(x))
M(V(b(x))) → V(Xb(x))
Xa(E(x)) → a(E(x))
Xb(E(x)) → b(E(x))
W(V(x)) → R(L(x))
L(a(x)) → D(b(x))
R(D(x)) → B(x)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Q is empty.
(3) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
XA(a(x)) → XA(x)
XA(b(x)) → XA(x)
XB(a(x)) → XB(x)
XB(b(x)) → XB(x)
L1(a(x)) → YA(L(x))
L1(a(x)) → L1(x)
L1(b(x)) → YB(L(x))
L1(b(x)) → L1(x)
The TRS R consists of the following rules:
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
The TRS R consists of the following rules:
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
R is empty.
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
We have to consider all minimal (P,Q,R)-chains.
(12) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- L1(b(x)) → L1(x)
The graph contains the following edges 1 > 1
- L1(a(x)) → L1(x)
The graph contains the following edges 1 > 1
(15) YES
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
XB(b(x)) → XB(x)
XB(a(x)) → XB(x)
The TRS R consists of the following rules:
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
XB(b(x)) → XB(x)
XB(a(x)) → XB(x)
R is empty.
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
XB(b(x)) → XB(x)
XB(a(x)) → XB(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- XB(b(x)) → XB(x)
The graph contains the following edges 1 > 1
- XB(a(x)) → XB(x)
The graph contains the following edges 1 > 1
(22) YES
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
XA(b(x)) → XA(x)
XA(a(x)) → XA(x)
The TRS R consists of the following rules:
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
We have to consider all minimal (P,Q,R)-chains.
(24) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
XA(b(x)) → XA(x)
XA(a(x)) → XA(x)
R is empty.
The set Q consists of the following terms:
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
We have to consider all minimal (P,Q,R)-chains.
(26) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
Xa(a(x0))
Xa(b(x0))
Xb(a(x0))
Xb(b(x0))
L(a(x0))
L(b(x0))
Ya(D(x0))
Yb(D(x0))
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
XA(b(x)) → XA(x)
XA(a(x)) → XA(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- XA(b(x)) → XA(x)
The graph contains the following edges 1 > 1
- XA(a(x)) → XA(x)
The graph contains the following edges 1 > 1
(29) YES