(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(.(x1, x2)) = x1 + x2
POL(f(x1)) = 1 + x1
POL(g(x1)) = 1 + x1
POL(nil) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(nil) → nil
g(nil) → nil
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(.(x1, x2)) = x1 + x2
POL(f(x1)) = 2·x1
POL(g(x1)) = 2·x1
POL(nil) = 2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(.(nil, y)) → .(nil, f(y))
g(.(x, nil)) → .(g(x), nil)
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
(5) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The signature Sigma is {
f,
g}
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, .(y, z))) → G(.(.(x, y), z))
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(10) Complex Obligation (AND)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(.(x, .(y, z))) → G(.(.(x, y), z))
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(12) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(.(x, .(y, z))) → G(.(.(x, y), z))
R is empty.
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(14) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(.(x, .(y, z))) → G(.(.(x, y), z))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) 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:
G(.(x, .(y, z))) → G(.(.(x, y), z))
Used ordering: Polynomial interpretation [POLO]:
POL(.(x1, x2)) = 2 + x1 + 2·x2
POL(G(x1)) = 2·x1
(17) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(19) TRUE
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(.(.(x, y), z)) → F(.(x, .(y, z)))
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(21) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(.(.(x, y), z)) → F(.(x, .(y, z)))
R is empty.
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(23) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(.(.(x, y), z)) → F(.(x, .(y, z)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) 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:
F(.(.(x, y), z)) → F(.(x, .(y, z)))
Used ordering: Polynomial interpretation [POLO]:
POL(.(x1, x2)) = 2 + 2·x1 + x2
POL(F(x1)) = 2·x1
(26) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(28) TRUE