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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

norm(nil) → 0
norm(g(z0, z1)) → s(norm(z0))
f(z0, nil) → g(nil, z0)
f(z0, g(z1, z2)) → g(f(z0, z1), z2)
rem(nil, z0) → nil
rem(g(z0, z1), 0) → g(z0, z1)
rem(g(z0, z1), s(z2)) → rem(z0, z2)

Tuples:

NORM(nil) → c
NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, nil) → c2
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(nil, z0) → c4
REM(g(z0, z1), 0) → c5
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

S tuples:

NORM(nil) → c
NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, nil) → c2
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(nil, z0) → c4
REM(g(z0, z1), 0) → c5
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

K tuples:none
Defined Rule Symbols:

norm, f, rem

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c, c1, c2, c3, c4, c5, c6

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 4 trailing nodes:

NORM(nil) → c
REM(g(z0, z1), 0) → c5
F(z0, nil) → c2
REM(nil, z0) → c4

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

norm(nil) → 0
norm(g(z0, z1)) → s(norm(z0))
f(z0, nil) → g(nil, z0)
f(z0, g(z1, z2)) → g(f(z0, z1), z2)
rem(nil, z0) → nil
rem(g(z0, z1), 0) → g(z0, z1)
rem(g(z0, z1), s(z2)) → rem(z0, z2)

Tuples:

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

S tuples:

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

K tuples:none
Defined Rule Symbols:

norm, f, rem

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c1, c3, c6

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

norm(nil) → 0
norm(g(z0, z1)) → s(norm(z0))
f(z0, nil) → g(nil, z0)
f(z0, g(z1, z2)) → g(f(z0, z1), z2)
rem(nil, z0) → nil
rem(g(z0, z1), 0) → g(z0, z1)
rem(g(z0, z1), s(z2)) → rem(z0, z2)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

S tuples:

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c1, c3, c6

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

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

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

We considered the (Usable) Rules:none
And the Tuples:

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

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

POL(F(x₁, x₂)) = x₂   
POL(NORM(x₁)) = [2]x₁   
POL(REM(x₁, x₂)) = [3]x₂   
POL(c1(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(g(x₁, x₂)) = [1] + x₁   
POL(s(x₁)) = [1] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

S tuples:none
K tuples:

NORM(g(z0, z1)) → c1(NORM(z0))
F(z0, g(z1, z2)) → c3(F(z0, z1))
REM(g(z0, z1), s(z2)) → c6(REM(z0, z2))

Defined Rule Symbols:none

Defined Pair Symbols:

NORM, F, REM

Compound Symbols:

c1, c3, c6

(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

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