(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(a, x) → g1(x, x)
f1(x, a) → g2(x, x)
f2(a, x) → g1(x, x)
f2(x, a) → g2(x, x)
g1(a, x) → h1(x)
g1(x, a) → h2(x)
g2(a, x) → h1(x)
g2(x, a) → h2(x)
h1(a) → i
h2(a) → i
e1(h1(w), h2(w), x, y, z, w) → e2(x, x, y, z, z, w)
e1(x1, x1, x, y, z, a) → e5(x1, x, y, z)
e2(f1(w, w), x, y, z, f2(w, w), w) → e3(x, y, x, y, y, z, y, z, x, y, z, w)
e2(x, x, y, z, z, a) → e6(x, y, z)
e2(i, x, y, z, i, a) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w)
e3(x, y, x, y, y, z, y, z, x, y, z, a) → e6(x, y, z)
e4(g1(w, w), x1, g2(w, w), x1, g1(w, w), x1, g2(w, w), x1, x, y, z, w) → e1(x1, x1, x, y, z, w)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z, a) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x, a) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)
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:
F1(a, x) → G1(x, x)
F1(x, a) → G2(x, x)
F2(a, x) → G1(x, x)
F2(x, a) → G2(x, x)
G1(a, x) → H1(x)
G1(x, a) → H2(x)
G2(a, x) → H1(x)
G2(x, a) → H2(x)
E1(h1(w), h2(w), x, y, z, w) → E2(x, x, y, z, z, w)
E1(x1, x1, x, y, z, a) → E5(x1, x, y, z)
E2(f1(w, w), x, y, z, f2(w, w), w) → E3(x, y, x, y, y, z, y, z, x, y, z, w)
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w) → E4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w)
E4(g1(w, w), x1, g2(w, w), x1, g1(w, w), x1, g2(w, w), x1, x, y, z, w) → E1(x1, x1, x, y, z, w)
E4(i, x1, i, x1, i, x1, i, x1, x, y, z, a) → E5(x1, x, y, z)
The TRS R consists of the following rules:
f1(a, x) → g1(x, x)
f1(x, a) → g2(x, x)
f2(a, x) → g1(x, x)
f2(x, a) → g2(x, x)
g1(a, x) → h1(x)
g1(x, a) → h2(x)
g2(a, x) → h1(x)
g2(x, a) → h2(x)
h1(a) → i
h2(a) → i
e1(h1(w), h2(w), x, y, z, w) → e2(x, x, y, z, z, w)
e1(x1, x1, x, y, z, a) → e5(x1, x, y, z)
e2(f1(w, w), x, y, z, f2(w, w), w) → e3(x, y, x, y, y, z, y, z, x, y, z, w)
e2(x, x, y, z, z, a) → e6(x, y, z)
e2(i, x, y, z, i, a) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w)
e3(x, y, x, y, y, z, y, z, x, y, z, a) → e6(x, y, z)
e4(g1(w, w), x1, g2(w, w), x1, g1(w, w), x1, g2(w, w), x1, x, y, z, w) → e1(x1, x1, x, y, z, w)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z, a) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x, a) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)
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 1 SCC with 10 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
E2(f1(w, w), x, y, z, f2(w, w), w) → E3(x, y, x, y, y, z, y, z, x, y, z, w)
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w) → E4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w)
E4(g1(w, w), x1, g2(w, w), x1, g1(w, w), x1, g2(w, w), x1, x, y, z, w) → E1(x1, x1, x, y, z, w)
E1(h1(w), h2(w), x, y, z, w) → E2(x, x, y, z, z, w)
The TRS R consists of the following rules:
f1(a, x) → g1(x, x)
f1(x, a) → g2(x, x)
f2(a, x) → g1(x, x)
f2(x, a) → g2(x, x)
g1(a, x) → h1(x)
g1(x, a) → h2(x)
g2(a, x) → h1(x)
g2(x, a) → h2(x)
h1(a) → i
h2(a) → i
e1(h1(w), h2(w), x, y, z, w) → e2(x, x, y, z, z, w)
e1(x1, x1, x, y, z, a) → e5(x1, x, y, z)
e2(f1(w, w), x, y, z, f2(w, w), w) → e3(x, y, x, y, y, z, y, z, x, y, z, w)
e2(x, x, y, z, z, a) → e6(x, y, z)
e2(i, x, y, z, i, a) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z, w)
e3(x, y, x, y, y, z, y, z, x, y, z, a) → e6(x, y, z)
e4(g1(w, w), x1, g2(w, w), x1, g1(w, w), x1, g2(w, w), x1, x, y, z, w) → e1(x1, x1, x, y, z, w)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z, a) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x, a) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.