YES Termination proof of rtL-me3.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRS S Cleaner (⇔, 0 ms)
↳2 RelTRS
↳3 RelTRSRRRProof (⇔, 83 ms)
↳4 RelTRS
↳5 RelTRSRRRProof (⇔, 13 ms)
↳6 RelTRS
↳7 RelTRSRRRProof (⇔, 0 ms)
↳8 RelTRS
↳9 RIsEmptyProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

topB(i, N1(x), y) → topA(1, T1(x), y)
topA(i, x, N2(y)) → topB(0, x, T2(y))
topB(i, S1(x), y) → topA(i, N1(x), y)
topA(i, x, S2(y)) → topB(i, x, N2(y))
topA(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

topB(i, x, N2(y)) → topB(i, x, N2(C(y)))
topA(i, S1(x), y) → topA(i, N1(x), y)
topB(i, x, S2(y)) → topB(i, x, N2(y))
topB(i, x, N2(y)) → topB(0, x, T2(y))
topA(i, T1(x), y) → topA(i, T1(x), y)
topB(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))
topB(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(i, N1(x), y) → topA(i, N1(C(x)), y)
topB(i, x, T2(y)) → topB(i, x, T2(y))
topB(i, x, S2(y)) → topB(i, x, S2(D(y)))
topA(i, N1(x), y) → topA(1, T1(x), y)

(1) RelTRS S Cleaner (EQUIVALENT transformation)

We have deleted all rules from S that have the shape t → t:

topA(i, T1(x), y) → topA(i, T1(x), y)
topB(i, x, T2(y)) → topB(i, x, T2(y))

(2) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

topB(i, N1(x), y) → topA(1, T1(x), y)
topA(i, x, N2(y)) → topB(0, x, T2(y))
topB(i, S1(x), y) → topA(i, N1(x), y)
topA(i, x, S2(y)) → topB(i, x, N2(y))
topA(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

topB(i, x, N2(y)) → topB(i, x, N2(C(y)))
topA(i, S1(x), y) → topA(i, N1(x), y)
topB(i, x, S2(y)) → topB(i, x, N2(y))
topB(i, x, N2(y)) → topB(0, x, T2(y))
topB(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))
topB(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(i, N1(x), y) → topA(i, N1(C(x)), y)
topB(i, x, S2(y)) → topB(i, x, S2(D(y)))
topA(i, N1(x), y) → topA(1, T1(x), y)

(3) RelTRSRRRProof (EQUIVALENT transformation)

We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(1) = 0   
POL(C(x1)) = x1   
POL(D(x1)) = x1   
POL(N1(x1)) = 1 + x1   
POL(N2(x1)) = 1 + x1   
POL(S1(x1)) = 1 + x1   
POL(S2(x1)) = 1 + x1   
POL(T1(x1)) = x1   
POL(T2(x1)) = 1 + x1   
POL(topA(x1, x2, x3)) = x1 + x2 + x3   
POL(topB(x1, x2, x3)) = x1 + x2 + x3   
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

topB(i, N1(x), y) → topA(1, T1(x), y)
Rules from S:

topA(i, N1(x), y) → topA(1, T1(x), y)


(4) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

topA(i, x, N2(y)) → topB(0, x, T2(y))
topB(i, S1(x), y) → topA(i, N1(x), y)
topA(i, x, S2(y)) → topB(i, x, N2(y))
topA(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

topB(i, x, N2(y)) → topB(i, x, N2(C(y)))
topA(i, S1(x), y) → topA(i, N1(x), y)
topB(i, x, S2(y)) → topB(i, x, N2(y))
topB(i, x, N2(y)) → topB(0, x, T2(y))
topB(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))
topB(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(i, N1(x), y) → topA(i, N1(C(x)), y)
topB(i, x, S2(y)) → topB(i, x, S2(D(y)))

(5) RelTRSRRRProof (EQUIVALENT transformation)

We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(1) = 0   
POL(C(x1)) = x1   
POL(D(x1)) = x1   
POL(N1(x1)) = x1   
POL(N2(x1)) = 1 + x1   
POL(S1(x1)) = 1 + x1   
POL(S2(x1)) = 1 + x1   
POL(T1(x1)) = x1   
POL(T2(x1)) = 1 + x1   
POL(topA(x1, x2, x3)) = x1 + x2 + x3   
POL(topB(x1, x2, x3)) = x1 + x2 + x3   
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

topB(i, S1(x), y) → topA(i, N1(x), y)
Rules from S:

topA(i, S1(x), y) → topA(i, N1(x), y)


(6) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

topA(i, x, N2(y)) → topB(0, x, T2(y))
topA(i, x, S2(y)) → topB(i, x, N2(y))
topA(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

topB(i, x, N2(y)) → topB(i, x, N2(C(y)))
topB(i, x, S2(y)) → topB(i, x, N2(y))
topB(i, x, N2(y)) → topB(0, x, T2(y))
topB(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))
topB(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(i, N1(x), y) → topA(i, N1(C(x)), y)
topB(i, x, S2(y)) → topB(i, x, S2(D(y)))

(7) RelTRSRRRProof (EQUIVALENT transformation)

We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(1) = 0   
POL(C(x1)) = x1   
POL(D(x1)) = x1   
POL(N1(x1)) = x1   
POL(N2(x1)) = x1   
POL(S2(x1)) = x1   
POL(T1(x1)) = x1   
POL(T2(x1)) = x1   
POL(topA(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(topB(x1, x2, x3)) = x1 + x2 + x3   
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

topA(i, x, N2(y)) → topB(0, x, T2(y))
topA(i, x, S2(y)) → topB(i, x, N2(y))
topA(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))
Rules from S:
none


(8) Obligation:

Relative term rewrite system:
R is empty.
The relative TRS consists of the following S rules:

topB(i, x, N2(y)) → topB(i, x, N2(C(y)))
topB(i, x, S2(y)) → topB(i, x, N2(y))
topB(i, x, N2(y)) → topB(0, x, T2(y))
topB(1, T1(x), T2(y)) → topB(1, T1(x), S2(y))
topB(i, N1(x), T2(y)) → topB(i, N1(x), S2(y))
topA(i, N1(x), y) → topA(i, N1(C(x)), y)
topB(i, x, S2(y)) → topB(i, x, S2(D(y)))

(9) RIsEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(10) YES