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:

f(g(i(a, b, b'), c), d) → if(e, f(.(b, c), d'), f(.(b', c), d'))
f(g(h(a, b), c), d) → if(e, f(.(b, g(h(a, b), c)), d), f(c, d'))

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(g(i(a, b, b'), c), d) → if(e, f(.(b, c), d'), f(.(b', c), d'))
f(g(h(a, b), c), d) → if(e, f(.(b, g(h(a, b), c)), d), f(c, d'))

Q is empty.

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

f(g(i(a, b, b'), c), d) → if(e, f(.(b, c), d'), f(.(b', c), d'))
f(g(h(a, b), c), d) → if(e, f(.(b, g(h(a, b), c)), d), f(c, d'))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

f(g(i(a, b, b'), c), d) → if(e, f(.(b, c), d'), f(.(b', c), d'))
Used ordering:
Polynomial interpretation [25]:

POL(.(x1, x2)) = x1 + 2·x2   
POL(a) = 0   
POL(b) = 0   
POL(b') = 1   
POL(c) = 0   
POL(d) = 0   
POL(d') = 0   
POL(e) = 0   
POL(f(x1, x2)) = x1 + 2·x2   
POL(g(x1, x2)) = 2·x1 + x2   
POL(h(x1, x2)) = 2·x1 + x2   
POL(i(x1, x2, x3)) = 2·x1 + x2 + 2·x3   
POL(if(x1, x2, x3)) = x1 + 2·x2 + 2·x3   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence

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

f(g(h(a, b), c), d) → if(e, f(.(b, g(h(a, b), c)), d), f(c, d'))

Q is empty.

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof

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

f(g(h(a, b), c), d) → if(e, f(.(b, g(h(a, b), c)), d), f(c, d'))

The set Q consists of the following terms:

f(g(h(a, b), c), d)


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

F(g(h(a, b), c), d) → F(c, d')
F(g(h(a, b), c), d) → F(.(b, g(h(a, b), c)), d)

The TRS R consists of the following rules:

f(g(h(a, b), c), d) → if(e, f(.(b, g(h(a, b), c)), d), f(c, d'))

The set Q consists of the following terms:

f(g(h(a, b), c), d)

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

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

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

F(g(h(a, b), c), d) → F(c, d')
F(g(h(a, b), c), d) → F(.(b, g(h(a, b), c)), d)

The TRS R consists of the following rules:

f(g(h(a, b), c), d) → if(e, f(.(b, g(h(a, b), c)), d), f(c, d'))

The set Q consists of the following terms:

f(g(h(a, b), c), d)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.