The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

s(a) → a
s(s(z0)) → z0
s(f(z0, z1)) → f(s(z1), s(z0))
s(g(z0, z1)) → g(s(z0), s(z1))
f(z0, a) → z0
f(a, z0) → z0
f(g(z0, z1), g(z2, z3)) → g(f(z0, z2), f(z1, z3))
g(a, a) → a

Tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

S tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

K tuples:none
Defined Rule Symbols:

s, f, g

Defined Pair Symbols:

S, F

Compound Symbols:

c2, c3, c6

(3) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))

We considered the (Usable) Rules:

f(z0, a) → z0
f(a, z0) → z0
f(g(z0, z1), g(z2, z3)) → g(f(z0, z2), f(z1, z3))
g(a, a) → a
s(a) → a
s(s(z0)) → z0
s(f(z0, z1)) → f(s(z1), s(z0))
s(g(z0, z1)) → g(s(z0), s(z1))

And the Tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x₁, x₂)) = 0   
POL(G(x₁, x₂)) = 0   
POL(S(x₁)) = [2] + [2]x₁   
POL(a) = [4]   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(f(x₁, x₂)) = [1] + [5]x₁ + [2]x₂   
POL(g(x₁, x₂)) = [5] + [5]x₁ + [3]x₂   
POL(s(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

s(a) → a
s(s(z0)) → z0
s(f(z0, z1)) → f(s(z1), s(z0))
s(g(z0, z1)) → g(s(z0), s(z1))
f(z0, a) → z0
f(a, z0) → z0
f(g(z0, z1), g(z2, z3)) → g(f(z0, z2), f(z1, z3))
g(a, a) → a

Tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

S tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

K tuples:

S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))

Defined Rule Symbols:

s, f, g

Defined Pair Symbols:

S, F

Compound Symbols:

c2, c3, c6

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))

We considered the (Usable) Rules:

f(z0, a) → z0
f(a, z0) → z0
f(g(z0, z1), g(z2, z3)) → g(f(z0, z2), f(z1, z3))
g(a, a) → a
s(a) → a
s(s(z0)) → z0
s(f(z0, z1)) → f(s(z1), s(z0))
s(g(z0, z1)) → g(s(z0), s(z1))

And the Tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x₁, x₂)) = 0   
POL(G(x₁, x₂)) = 0   
POL(S(x₁)) = [1] + [2]x₁   
POL(a) = [1]   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(f(x₁, x₂)) = [1] + [3]x₁ + [4]x₂   
POL(g(x₁, x₂)) = [5] + [5]x₁ + [3]x₂   
POL(s(x₁)) = [3]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

s(a) → a
s(s(z0)) → z0
s(f(z0, z1)) → f(s(z1), s(z0))
s(g(z0, z1)) → g(s(z0), s(z1))
f(z0, a) → z0
f(a, z0) → z0
f(g(z0, z1), g(z2, z3)) → g(f(z0, z2), f(z1, z3))
g(a, a) → a

Tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

S tuples:

F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

K tuples:

S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))

Defined Rule Symbols:

s, f, g

Defined Pair Symbols:

S, F

Compound Symbols:

c2, c3, c6

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

We considered the (Usable) Rules:

f(z0, a) → z0
f(a, z0) → z0
f(g(z0, z1), g(z2, z3)) → g(f(z0, z2), f(z1, z3))
g(a, a) → a
s(a) → a
s(s(z0)) → z0
s(f(z0, z1)) → f(s(z1), s(z0))
s(g(z0, z1)) → g(s(z0), s(z1))

And the Tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x₁, x₂)) = [3] + [4]x₁ + [4]x₂   
POL(G(x₁, x₂)) = [2] + x₁ + x₂   
POL(S(x₁)) = [2]x₁   
POL(a) = [1]   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c3(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c6(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(f(x₁, x₂)) = [3] + [3]x₁ + [3]x₂   
POL(g(x₁, x₂)) = [4] + [4]x₁ + [2]x₂   
POL(s(x₁)) = x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

s(a) → a
s(s(z0)) → z0
s(f(z0, z1)) → f(s(z1), s(z0))
s(g(z0, z1)) → g(s(z0), s(z1))
f(z0, a) → z0
f(a, z0) → z0
f(g(z0, z1), g(z2, z3)) → g(f(z0, z2), f(z1, z3))
g(a, a) → a

Tuples:

S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

S tuples:none
K tuples:

S(g(z0, z1)) → c3(G(s(z0), s(z1)), S(z0), S(z1))
S(f(z0, z1)) → c2(F(s(z1), s(z0)), S(z1), S(z0))
F(g(z0, z1), g(z2, z3)) → c6(G(f(z0, z2), f(z1, z3)), F(z0, z2), F(z1, z3))

Defined Rule Symbols:

s, f, g

Defined Pair Symbols:

S, F

Compound Symbols:

c2, c3, c6

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))