### (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