(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(x, plus(y, s(z))) → PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) → PLUS(y, times(s(z), 0))
TIMES(x, plus(y, s(z))) → TIMES(s(z), 0)
TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, s(y)) → TIMES(x, y)
PLUS(x, s(y)) → PLUS(x, y)
The TRS R consists of the following rules:
times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(x, y)
The TRS R consists of the following rules:
times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- PLUS(x, s(y)) → PLUS(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
The TRS R consists of the following rules:
times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
The TRS R consists of the following rules:
times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)
Strictly oriented rules of the TRS R:
plus(x, 0) → x
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(TIMES(x1, x2)) = x1 + 2·x2
POL(plus(x1, x2)) = 1 + x1 + x2
POL(s(x1)) = 1 + x1
POL(times(x1, x2)) = x1 + x2
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
The TRS R consists of the following rules:
times(x, 0) → 0
plus(x, s(y)) → s(plus(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
plus(x, s(y)) → s(plus(x, y))
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(TIMES(x1, x2)) = x1 + 2·x2
POL(plus(x1, x2)) = x1 + 2·x2
POL(s(x1)) = 1 + x1
POL(times(x1, x2)) = x1 + 2·x2
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
The TRS R consists of the following rules:
times(x, 0) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(18) TRUE