(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))
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:
PLUS(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))
PLUS(s(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X3, plus(X2, X4))
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X2, X4)
The TRS R consists of the following rules:
plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) 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:
PLUS(s(X), plus(Y, Z)) → PLUS(s(s(Y)), Z)
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X3, plus(X2, X4))
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X2, X4)
Used ordering: Polynomial interpretation [POLO]:
POL(PLUS(x1, x2)) = x1 + x2
POL(plus(x1, x2)) = 1 + x1 + x2
POL(s(x1)) = x1
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))
PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
The TRS R consists of the following rules:
plus(s(X), plus(Y, Z)) → plus(X, plus(s(s(Y)), Z))
plus(s(X1), plus(X2, plus(X3, X4))) → plus(X1, plus(X3, plus(X2, X4)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) 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(s(X), plus(Y, Z)) → PLUS(X, plus(s(s(Y)), Z))
The graph contains the following edges 1 > 1
- PLUS(s(X1), plus(X2, plus(X3, X4))) → PLUS(X1, plus(X3, plus(X2, X4)))
The graph contains the following edges 1 > 1
(6) TRUE