Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(f(x₁, x₂)) = 1 + x₁ + x₂   
POL(g(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

g(0, f(x, x)) → x

(2) Obligation:

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

g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(f(x₁, x₂)) = x₁ + x₂   
POL(g(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)

(4) Obligation:

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

g(f(x, y), 0) → f(g(x, 0), g(y, 0))

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(f(x₁, x₂)) = 1 + 2·x₁ + x₂   
POL(g(x₁, x₂)) = 2·x₁ + x₂   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

g(f(x, y), 0) → f(g(x, 0), g(y, 0))

(6) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(7) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(8) TRUE

(9) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(10) TRUE