YES
0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPReductionPairProof (⇔, 58 ms)
↳4 RelADPP
↳5 RelADPReductionPairProof (⇔, 0 ms)
↳6 RelADPP
↳7 DAbsisEmptyProof (⇔, 0 ms)
↳8 YES
f(g(x), y, z) → f(x, y, g(z))
f(x, y, z) → f(x, y, g(z))
f(x, a, z) → f(x, g(a), z)
f(x, y, g(z)) → f(x, g(y), z)
We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].
f(g(x), y, z) → F(x, y, g(z))
f(x, y, z) → F(x, y, g(z))
f(x, a, z) → F(x, g(a), z)
f(x, y, g(z)) → F(x, g(y), z)
We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Relative ADPs:
f(x, a, z) → F(x, g(a), z)
f(g(x), y, z) → F(x, y, g(z))
f(x, y, g(z)) → F(x, g(y), z)
f(x, y, z) → F(x, y, g(z))
POL(F(x1, x2, x3)) = 2·x2
POL(a) = 1
POL(f(x1, x2, x3)) = 2·x2
POL(g(x1)) = 0
f(g(x), y, z) → F(x, y, g(z))
f(x, y, g(z)) → F(x, g(y), z)
f(x, y, z) → F(x, y, g(z))
f(x, a, z) → f(x, g(a), z)
We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:
f(g(x), y, z) → F(x, y, g(z))
f(x, a, z) → f(x, g(a), z)
POL(F(x1, x2, x3)) = 2 + 3·x1
POL(a) = 0
POL(f(x1, x2, x3)) = 0
POL(g(x1)) = 2 + x1
f(x, y, g(z)) → F(x, g(y), z)
f(x, y, z) → F(x, y, g(z))
f(x, a, z) → f(x, g(a), z)
f(g(x), y, z) → f(x, y, g(z))