YES
proof of Transformed_CSR_04_PEANO_nosorts_iGM.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 68 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 20 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) RisEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(and(tt, X)) -> mark(X)
   active(plus(N, 0)) -> mark(N)
   active(plus(N, s(M))) -> mark(s(plus(N, M)))
   mark(and(X1, X2)) -> active(and(mark(X1), X2))
   mark(tt) -> active(tt)
   mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
   mark(0) -> active(0)
   mark(s(X)) -> active(s(mark(X)))
   and(mark(X1), X2) -> and(X1, X2)
   and(X1, mark(X2)) -> and(X1, X2)
   and(active(X1), X2) -> and(X1, X2)
   and(X1, active(X2)) -> and(X1, X2)
   plus(mark(X1), X2) -> plus(X1, X2)
   plus(X1, mark(X2)) -> plus(X1, X2)
   plus(active(X1), X2) -> plus(X1, X2)
   plus(X1, active(X2)) -> plus(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(active(x_1)) = x_1
   POL(and(x_1, x_2)) = 1 + x_1 + x_2
   POL(mark(x_1)) = x_1
   POL(plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
   POL(s(x_1)) = 1 + x_1
   POL(tt) = 1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   active(and(tt, X)) -> mark(X)
   active(plus(N, 0)) -> mark(N)
   active(plus(N, s(M))) -> mark(s(plus(N, M)))




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   mark(and(X1, X2)) -> active(and(mark(X1), X2))
   mark(tt) -> active(tt)
   mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
   mark(0) -> active(0)
   mark(s(X)) -> active(s(mark(X)))
   and(mark(X1), X2) -> and(X1, X2)
   and(X1, mark(X2)) -> and(X1, X2)
   and(active(X1), X2) -> and(X1, X2)
   and(X1, active(X2)) -> and(X1, X2)
   plus(mark(X1), X2) -> plus(X1, X2)
   plus(X1, mark(X2)) -> plus(X1, X2)
   plus(active(X1), X2) -> plus(X1, X2)
   plus(X1, active(X2)) -> plus(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 2
   POL(active(x_1)) = 1 + x_1
   POL(and(x_1, x_2)) = 2 + x_1 + x_2
   POL(mark(x_1)) = 1 + 2*x_1
   POL(plus(x_1, x_2)) = 2 + x_1 + x_2
   POL(s(x_1)) = 2 + x_1
   POL(tt) = 1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   mark(and(X1, X2)) -> active(and(mark(X1), X2))
   mark(tt) -> active(tt)
   mark(0) -> active(0)
   mark(s(X)) -> active(s(mark(X)))
   and(mark(X1), X2) -> and(X1, X2)
   and(X1, mark(X2)) -> and(X1, X2)
   and(active(X1), X2) -> and(X1, X2)
   and(X1, active(X2)) -> and(X1, X2)
   plus(mark(X1), X2) -> plus(X1, X2)
   plus(X1, mark(X2)) -> plus(X1, X2)
   plus(active(X1), X2) -> plus(X1, X2)
   plus(X1, active(X2)) -> plus(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)




----------------------------------------

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))

Q is empty.

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(5) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(active(x_1)) = x_1
   POL(mark(x_1)) = 2*x_1
   POL(plus(x_1, x_2)) = 1 + x_1 + 2*x_2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))




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(6)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(7) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(8)
YES