YES
0 QTRS
↳1 QTRSToCSRProof (⇔, 0 ms)
↳2 CSR
↳3 CSRRRRProof (⇔, 0 ms)
↳4 CSR
↳5 CSDependencyPairsProof (⇔, 0 ms)
↳6 QCSDP
↳7 QCSDependencyGraphProof (⇔, 0 ms)
↳8 TRUE
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
f: {1}
a: empty set
b: empty set
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).
f(X, X) → f(a, b)
b → a
f: {1}
a: empty set
b: empty set
f(X, X) → f(a, b)
b → a
f: {1}
a: empty set
b: empty set
Used ordering:
Polynomial interpretation [POLO]:
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
POL(a) = 0
POL(b) = 1
POL(f(x1, x2)) = x1
b → a
f(X, X) → f(a, b)
f: {1}
a: empty set
b: empty set
F(X, X) → F(a, b)
f(X, X) → f(a, b)