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, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
f: empty set
g: {1}
b: empty set
c: empty set
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).
f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c
f: empty set
g: {1}
b: empty set
c: empty set
f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c
f: empty set
g: {1}
b: empty set
c: 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(b) = 1
POL(c) = 0
POL(f(x1, x2, x3)) = 0
POL(g(x1)) = 2·x1
g(b) → c
b → c
f(X, g(X), Y) → f(Y, Y, Y)
f: empty set
g: {1}
F(X, g(X), Y) → F(Y, Y, Y)
f(X, g(X), Y) → f(Y, Y, Y)