Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF1(false, x, y, c, i, j) → SIZE(j)
REACH(x, y, c, i, j) → EQ(x, y)
IF1(false, x, y, c, i, j) → LE(c, size(j))
LE(s(x), s(y)) → LE(x, y)
IF1(false, x, y, c, i, j) → IF2(le(c, size(j)), x, y, c, i, j)
EQ(s(x), s(y)) → EQ(x, y)
REACHABLE(x, y, i) → REACH(x, y, 0, i, i)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → OR(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))
IF2(true, x, y, c, edge(u, v, i), j) → AND(eq(x, u), reach(v, y, s(c), j, j))
IF2(true, x, y, c, edge(u, v, i), j) → EQ(x, u)
REACH(x, y, c, i, j) → IF1(eq(x, y), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)
SIZE(edge(x, y, i)) → SIZE(i)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

IF1(false, x, y, c, i, j) → SIZE(j)
REACH(x, y, c, i, j) → EQ(x, y)
IF1(false, x, y, c, i, j) → LE(c, size(j))
LE(s(x), s(y)) → LE(x, y)
IF1(false, x, y, c, i, j) → IF2(le(c, size(j)), x, y, c, i, j)
EQ(s(x), s(y)) → EQ(x, y)
REACHABLE(x, y, i) → REACH(x, y, 0, i, i)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → OR(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))
IF2(true, x, y, c, edge(u, v, i), j) → AND(eq(x, u), reach(v, y, s(c), j, j))
IF2(true, x, y, c, edge(u, v, i), j) → EQ(x, u)
REACH(x, y, c, i, j) → IF1(eq(x, y), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)
SIZE(edge(x, y, i)) → SIZE(i)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 7 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SIZE(edge(x, y, i)) → SIZE(i)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SIZE(edge(x, y, i)) → SIZE(i)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SIZE(edge(x, y, i)) → SIZE(i)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

IF1(false, x, y, c, i, j) → IF2(le(c, size(j)), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
REACH(x, y, c, i, j) → IF1(eq(x, y), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

IF1(false, x, y, c, i, j) → IF2(le(c, size(j)), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
REACH(x, y, c, i, j) → IF1(eq(x, y), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

or(true, x0)
or(false, x0)
and(true, x0)
and(false, x0)
reachable(x0, x1, x2)
reach(x0, x1, x2, x3, x4)
if1(true, x0, x1, x2, x3, x4)
if1(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, empty, x3)
if2(true, x0, x1, x2, edge(x3, x4, x5), x6)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ Instantiation

Q DP problem:
The TRS P consists of the following rules:

IF1(false, x, y, c, i, j) → IF2(le(c, size(j)), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
REACH(x, y, c, i, j) → IF1(eq(x, y), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule REACH(x, y, c, i, j) → IF1(eq(x, y), x, y, c, i, j) we obtained the following new rules:

REACH(z4, z1, s(z2), z6, z6) → IF1(eq(z4, z1), z4, z1, s(z2), z6, z6)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
QDP
                            ↳ Instantiation

Q DP problem:
The TRS P consists of the following rules:

REACH(z4, z1, s(z2), z6, z6) → IF1(eq(z4, z1), z4, z1, s(z2), z6, z6)
IF1(false, x, y, c, i, j) → IF2(le(c, size(j)), x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF1(false, x, y, c, i, j) → IF2(le(c, size(j)), x, y, c, i, j) we obtained the following new rules:

IF1(false, z0, z1, s(z2), z3, z3) → IF2(le(s(z2), size(z3)), z0, z1, s(z2), z3, z3)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
QDP
                                ↳ NonInfProof

Q DP problem:
The TRS P consists of the following rules:

REACH(z4, z1, s(z2), z6, z6) → IF1(eq(z4, z1), z4, z1, s(z2), z6, z6)
IF1(false, z0, z1, s(z2), z3, z3) → IF2(le(s(z2), size(z3)), z0, z1, s(z2), z3, z3)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus [18] with the following steps:
Note that final constraints are written in bold face.


For Pair IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j) the following chains were created:




For Pair IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j) the following chains were created:




For Pair REACH(z4, z1, s(z2), z6, z6) → IF1(eq(z4, z1), z4, z1, s(z2), z6, z6) the following chains were created:




For Pair IF1(false, z0, z1, s(z2), z3, z3) → IF2(le(s(z2), size(z3)), z0, z1, s(z2), z3, z3) the following chains were created:




To summarize, we get the following constraints P for the following pairs.



The constraints for P> respective Pbound are constructed from P where we just replace every occurence of "t ≥ s" in P by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [18]:

POL(0) = 0   
POL(IF1(x1, x2, x3, x4, x5, x6)) = -1 - x1 - x4 + x5   
POL(IF2(x1, x2, x3, x4, x5, x6)) = -1 - x4 + x6   
POL(REACH(x1, x2, x3, x4, x5)) = -x3 + x4   
POL(c) = -1   
POL(edge(x1, x2, x3)) = 1 + x3   
POL(empty) = 0   
POL(eq(x1, x2)) = 0   
POL(false) = 0   
POL(le(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   
POL(size(x1)) = x1   
POL(true) = 1   

The following pairs are in P>:

REACH(z4, z1, s(z2), z6, z6) → IF1(eq(z4, z1), z4, z1, s(z2), z6, z6)
The following pairs are in Pbound:

IF1(false, z0, z1, s(z2), z3, z3) → IF2(le(s(z2), size(z3)), z0, z1, s(z2), z3, z3)
The following rules are usable:

falseeq(0, s(x))
trueeq(0, 0)
eq(x, y) → eq(s(x), s(y))
falseeq(s(x), 0)


↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
QDP
                                      ↳ DependencyGraphProof
                                    ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF1(false, z0, z1, s(z2), z3, z3) → IF2(le(s(z2), size(z3)), z0, z1, s(z2), z3, z3)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
                                    ↳ QDP
                                      ↳ DependencyGraphProof
QDP
                                          ↳ UsableRulesProof
                                    ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ UsableRulesProof
QDP
                                              ↳ QReductionProof
                                    ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
QDP
                                                  ↳ QDPSizeChangeProof
                                    ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
                                    ↳ QDP
QDP
                                      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

REACH(z4, z1, s(z2), z6, z6) → IF1(eq(z4, z1), z4, z1, s(z2), z6, z6)
IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)
IF2(true, x, y, c, edge(u, v, i), j) → REACH(v, y, s(c), j, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
                                    ↳ QDP
                                    ↳ QDP
                                      ↳ DependencyGraphProof
QDP
                                          ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)

The TRS R consists of the following rules:

size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
                                    ↳ QDP
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ UsableRulesProof
QDP
                                              ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)

R is empty.
The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
size(empty)
size(edge(x0, x1, x2))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Instantiation
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ NonInfProof
                                  ↳ AND
                                    ↳ QDP
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ UsableRulesProof
                                            ↳ QDP
                                              ↳ QReductionProof
QDP
                                                  ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

IF2(true, x, y, c, edge(u, v, i), j) → IF2(true, x, y, c, i, j)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: