(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
B(x) → W(M(M(M(M(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(a(x))) → D(b(b(b(x))))
L(b(b(b(b(b(x)))))) → D(a(a(a(x))))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
R(D(x)) → B(x)
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
B(x) → V(M(M(M(M(W(x))))))
M(x) → x
a(V(M(x))) → Xa(V(x))
b(V(M(x))) → Xb(V(x))
a(Xa(x)) → Xa(a(x))
b(Xa(x)) → Xa(b(x))
a(Xb(x)) → Xb(a(x))
b(Xb(x)) → Xb(b(x))
E(Xa(x)) → E(a(x))
E(Xb(x)) → E(b(x))
V(W(x)) → L(R(x))
a(L(x)) → L(Ya(x))
b(L(x)) → L(Yb(x))
a(a(L(x))) → b(b(b(D(x))))
b(b(b(b(b(L(x)))))) → a(a(a(D(x))))
D(Ya(x)) → a(D(x))
D(Yb(x)) → b(D(x))
D(R(x)) → B(x)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(B(x1)) = 10 + x1
POL(D(x1)) = 11 + 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)) = 97 + x1
POL(Xb(x1)) = 61 + x1
POL(Ya(x1)) = 96 + x1
POL(Yb(x1)) = 60 + x1
POL(a(x1)) = 96 + x1
POL(b(x1)) = 60 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
B(x) → V(M(M(M(M(W(x))))))
M(x) → x
a(V(M(x))) → Xa(V(x))
b(V(M(x))) → Xb(V(x))
E(Xa(x)) → E(a(x))
E(Xb(x)) → E(b(x))
V(W(x)) → L(R(x))
a(a(L(x))) → b(b(b(D(x))))
b(b(b(b(b(L(x)))))) → a(a(a(D(x))))
D(R(x)) → B(x)
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(Xa(x)) → Xa(a(x))
b(Xa(x)) → Xa(b(x))
a(Xb(x)) → Xb(a(x))
b(Xb(x)) → Xb(b(x))
a(L(x)) → L(Ya(x))
b(L(x)) → L(Yb(x))
D(Ya(x)) → a(D(x))
D(Yb(x)) → b(D(x))
Q is empty.
(5) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(Xa(x)) → Xa(a(x))
b(Xa(x)) → Xa(b(x))
a(Xb(x)) → Xb(a(x))
b(Xb(x)) → Xb(b(x))
a(L(x)) → L(Ya(x))
b(L(x)) → L(Yb(x))
D(Ya(x)) → a(D(x))
D(Yb(x)) → b(D(x))
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(Xa(x)) → A(x)
B(Xa(x)) → B(x)
A(Xb(x)) → A(x)
B(Xb(x)) → B(x)
D1(Ya(x)) → A(D(x))
D1(Ya(x)) → D1(x)
D1(Yb(x)) → B(D(x))
D1(Yb(x)) → D1(x)
The TRS R consists of the following rules:
a(Xa(x)) → Xa(a(x))
b(Xa(x)) → Xa(b(x))
a(Xb(x)) → Xb(a(x))
b(Xb(x)) → Xb(b(x))
a(L(x)) → L(Ya(x))
b(L(x)) → L(Yb(x))
D(Ya(x)) → a(D(x))
D(Yb(x)) → b(D(x))
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
(10) Complex Obligation (AND)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(Xb(x)) → B(x)
B(Xa(x)) → B(x)
The TRS R consists of the following rules:
a(Xa(x)) → Xa(a(x))
b(Xa(x)) → Xa(b(x))
a(Xb(x)) → Xb(a(x))
b(Xb(x)) → Xb(b(x))
a(L(x)) → L(Ya(x))
b(L(x)) → L(Yb(x))
D(Ya(x)) → a(D(x))
D(Yb(x)) → b(D(x))
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
We have to consider all minimal (P,Q,R)-chains.
(12) 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.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(Xb(x)) → B(x)
B(Xa(x)) → B(x)
R is empty.
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
We have to consider all minimal (P,Q,R)-chains.
(14) 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].
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(Xb(x)) → B(x)
B(Xa(x)) → B(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) 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:
- B(Xb(x)) → B(x)
The graph contains the following edges 1 > 1
- B(Xa(x)) → B(x)
The graph contains the following edges 1 > 1
(17) YES
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(Xb(x)) → A(x)
A(Xa(x)) → A(x)
The TRS R consists of the following rules:
a(Xa(x)) → Xa(a(x))
b(Xa(x)) → Xa(b(x))
a(Xb(x)) → Xb(a(x))
b(Xb(x)) → Xb(b(x))
a(L(x)) → L(Ya(x))
b(L(x)) → L(Yb(x))
D(Ya(x)) → a(D(x))
D(Yb(x)) → b(D(x))
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) 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.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(Xb(x)) → A(x)
A(Xa(x)) → A(x)
R is empty.
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
We have to consider all minimal (P,Q,R)-chains.
(21) 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].
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(Xb(x)) → A(x)
A(Xa(x)) → A(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) 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:
- A(Xb(x)) → A(x)
The graph contains the following edges 1 > 1
- A(Xa(x)) → A(x)
The graph contains the following edges 1 > 1
(24) YES
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D1(Yb(x)) → D1(x)
D1(Ya(x)) → D1(x)
The TRS R consists of the following rules:
a(Xa(x)) → Xa(a(x))
b(Xa(x)) → Xa(b(x))
a(Xb(x)) → Xb(a(x))
b(Xb(x)) → Xb(b(x))
a(L(x)) → L(Ya(x))
b(L(x)) → L(Yb(x))
D(Ya(x)) → a(D(x))
D(Yb(x)) → b(D(x))
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
We have to consider all minimal (P,Q,R)-chains.
(26) 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.
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D1(Yb(x)) → D1(x)
D1(Ya(x)) → D1(x)
R is empty.
The set Q consists of the following terms:
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
We have to consider all minimal (P,Q,R)-chains.
(28) 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].
a(Xa(x0))
b(Xa(x0))
a(Xb(x0))
b(Xb(x0))
a(L(x0))
b(L(x0))
D(Ya(x0))
D(Yb(x0))
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D1(Yb(x)) → D1(x)
D1(Ya(x)) → D1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(30) 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:
- D1(Yb(x)) → D1(x)
The graph contains the following edges 1 > 1
- D1(Ya(x)) → D1(x)
The graph contains the following edges 1 > 1
(31) YES