YES
0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPReductionPairProof (⇔, 0 ms)
↳4 RelADPP
↳5 RelADPReductionPairProof (⇔, 6 ms)
↳6 RelADPP
↳7 DAbsisEmptyProof (⇔, 0 ms)
↳8 YES
R(x, B2) → B2
W(x, B2) → B2
B1 → W(T, B1)
B1 → R(T, B1)
We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].
R(x, B2) → B2
W(x, B2) → B2
B1 → W1(T, B11)
B1 → R1(T, B11)
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:
R(x, B2) → B2
W(x, B2) → B2
B1 → W1(T, B11)
B1 → R1(T, B11)
POL(B1) = 0
POL(B11) = 0
POL(B2) = 1
POL(R(x1, x2)) = x2
POL(R1(x1, x2)) = x2
POL(T) = 0
POL(W(x1, x2)) = x2
POL(W1(x1, x2)) = 0
W(x, B2) → B2
B1 → W1(T, B11)
B1 → R1(T, B11)
R(x, B2) → B2
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:
W(x, B2) → B2
R(x, B2) → B2
POL(B1) = 0
POL(B11) = 0
POL(B2) = 1
POL(R(x1, x2)) = x2
POL(R1(x1, x2)) = 0
POL(T) = 1
POL(W(x1, x2)) = x2
POL(W1(x1, x2)) = x2
B1 → W1(T, B11)
W(x, B2) → B2
B1 → R1(T, B11)
R(x, B2) → B2