YES Termination proof of rtL-me2.trs

(0) Obligation:

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

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

The relative TRS consists of the following S rules:

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

(1) RelTRSRRRProof (EQUIVALENT transformation)

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

POL(0) = 0   
POL(1) = 0   
POL(N1) = 1   
POL(N2) = 1   
POL(S1) = 1   
POL(S2) = 1   
POL(T1) = 0   
POL(T2) = 1   
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, y) → topA(1, T1, y)
Rules from S:

topA(i, N1, y) → topA(1, T1, y)


(2) Obligation:

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

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

The relative TRS consists of the following S rules:

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

(3) RelTRSRRRProof (EQUIVALENT transformation)

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

POL(0) = 0   
POL(1) = 0   
POL(N1) = 0   
POL(N2) = 0   
POL(S1) = 1   
POL(S2) = 0   
POL(T1) = 0   
POL(T2) = 0   
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, y) → topA(i, N1, y)
Rules from S:

topA(i, S1, y) → topA(i, N1, y)


(4) Obligation:

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

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

The relative TRS consists of the following S rules:

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

(5) RelTRSRRRProof (EQUIVALENT transformation)

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

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


(6) Obligation:

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

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

(7) RIsEmptyProof (EQUIVALENT transformation)

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

(8) YES