### (0) Obligation:

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

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
tp(mark(z0)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
Tuples:

ACTIVE(f(z0)) → c1(F(f(z0)), F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
CHK(no(c)) → c3(ACTIVE(c))
MAT(f(z0), f(y)) → c4(F(mat(z0, y)), MAT(z0, y))
MAT(f(z0), c) → c5
F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
S tuples:

ACTIVE(f(z0)) → c1(F(f(z0)), F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
CHK(no(c)) → c3(ACTIVE(c))
MAT(f(z0), f(y)) → c4(F(mat(z0, y)), MAT(z0, y))
MAT(f(z0), c) → c5
F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
K tuples:none
Defined Rule Symbols:

active, chk, mat, f, tp

Defined Pair Symbols:

ACTIVE, CHK, MAT, F, TP

Compound Symbols:

c1, c2, c3, c4, c5, c6, c7, c8, c9

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

Removed 3 trailing nodes:

CHK(no(c)) → c3(ACTIVE(c))
MAT(f(z0), f(y)) → c4(F(mat(z0, y)), MAT(z0, y))
MAT(f(z0), c) → c5

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
tp(mark(z0)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
Tuples:

ACTIVE(f(z0)) → c1(F(f(z0)), F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
S tuples:

ACTIVE(f(z0)) → c1(F(f(z0)), F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)), MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0), F(f(f(f(f(f(f(f(f(f(X)))))))))), F(f(f(f(f(f(f(f(f(X))))))))), F(f(f(f(f(f(f(f(X)))))))), F(f(f(f(f(f(f(X))))))), F(f(f(f(f(f(X)))))), F(f(f(f(f(X))))), F(f(f(f(X)))), F(f(f(X))), F(f(X)), F(X))
K tuples:none
Defined Rule Symbols:

active, chk, mat, f, tp

Defined Pair Symbols:

ACTIVE, CHK, F, TP

Compound Symbols:

c1, c2, c6, c7, c8, c9

### (5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 23 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
tp(mark(z0)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
K tuples:none
Defined Rule Symbols:

active, chk, mat, f, tp

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c9

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

tp(mark(z0)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
K tuples:none
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c9

### (9) 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.

F(no(z0)) → c7(F(z0))
We considered the (Usable) Rules:

f(no(z0)) → no(f(z0))
active(f(z0)) → mark(f(f(z0)))
f(active(z0)) → active(f(z0))
f(mark(z0)) → mark(f(z0))
chk(no(c)) → active(c)
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
And the Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVE(x1)) = [4]x1
POL(CHK(x1)) = 0
POL(F(x1)) = [4]x1
POL(TP(x1)) = 0
POL(X) = 0
POL(active(x1)) = [5]x1
POL(c) = 0
POL(c1(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1, x2)) = x1 + x2
POL(chk(x1)) = 0
POL(f(x1)) = [4]x1
POL(mark(x1)) = x1
POL(mat(x1, x2)) = 0
POL(no(x1)) = [4] + x1
POL(y) = 0

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
K tuples:

F(no(z0)) → c7(F(z0))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c9

### (11) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace CHK(no(f(z0))) → c2(F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))) by

CHK(no(f(f(y)))) → c2(F(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
CHK(no(f(f(y)))) → c2(F(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
CHK(no(f(f(y)))) → c2(F(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
K tuples:

F(no(z0)) → c7(F(z0))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, TP, CHK

Compound Symbols:

c6, c7, c8, c1, c9, c2

### (13) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
K tuples:

F(no(z0)) → c7(F(z0))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, TP, CHK

Compound Symbols:

c6, c7, c8, c1, c9, c2, c2

### (15) 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.

CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
We considered the (Usable) Rules:

mat(f(z0), c) → no(c)
mat(f(z0), f(y)) → f(mat(z0, y))
And the Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVE(x1)) = 0
POL(CHK(x1)) = [4]x1
POL(F(x1)) = 0
POL(TP(x1)) = 0
POL(X) = 0
POL(active(x1)) = 0
POL(c) = 0
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1, x2)) = x1 + x2
POL(chk(x1)) = 0
POL(f(x1)) = [2]x1
POL(mark(x1)) = 0
POL(mat(x1, x2)) = 0
POL(no(x1)) = x1
POL(y) = [2]

### (16) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, TP, CHK

Compound Symbols:

c6, c7, c8, c1, c9, c2, c2

### (17) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace TP(mark(z0)) → c9(TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0))) by

TP(mark(f(y))) → c9(TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))

### (18) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(f(y))) → c9(TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c2, c9

### (19) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

### (20) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
mat(f(z0), c) → no(c)
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c2, c9, c9

### (21) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

chk(no(f(z0))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), z0)))

### (22) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(c)) → active(c)
mat(f(z0), c) → no(c)
mat(f(z0), f(y)) → f(mat(z0, y))
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c2, c9, c9

### (23) 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.

TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
We considered the (Usable) Rules:

chk(no(c)) → active(c)
And the Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVE(x1)) = 0
POL(CHK(x1)) = 0
POL(F(x1)) = 0
POL(TP(x1)) = [1] + [2]x1
POL(X) = [2]
POL(active(x1)) = 0
POL(c) = [3]
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1)) = x1
POL(c9(x1, x2)) = x1 + x2
POL(chk(x1)) = [3]x1
POL(f(x1)) = [4]x1
POL(mark(x1)) = 0
POL(mat(x1, x2)) = [4] + [5]x1 + [2]x2
POL(no(x1)) = 0
POL(y) = 0

### (24) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(c)) → active(c)
mat(f(z0), c) → no(c)
mat(f(z0), f(y)) → f(mat(z0, y))
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c2, c9, c9

### (25) 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.

TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
We considered the (Usable) Rules:

chk(no(c)) → active(c)
And the Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVE(x1)) = 0
POL(CHK(x1)) = 0
POL(F(x1)) = 0
POL(TP(x1)) = [1] + [4]x1
POL(X) = [3]
POL(active(x1)) = [3] + [3]x1
POL(c) = 0
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1)) = x1
POL(c9(x1, x2)) = x1 + x2
POL(chk(x1)) = [4] + x1
POL(f(x1)) = [2] + [2]x1
POL(mark(x1)) = [5]
POL(mat(x1, x2)) = [5] + [2]x1 + [4]x2
POL(no(x1)) = 0
POL(y) = 0

### (26) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(c)) → active(c)
mat(f(z0), c) → no(c)
mat(f(z0), f(y)) → f(mat(z0, y))
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c2, c9, c9

### (27) 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.

CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
We considered the (Usable) Rules:

mat(f(z0), c) → no(c)
mat(f(z0), f(y)) → f(mat(z0, y))
And the Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVE(x1)) = 0
POL(CHK(x1)) = x1
POL(F(x1)) = 0
POL(TP(x1)) = [1]
POL(X) = 0
POL(active(x1)) = 0
POL(c) = 0
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1)) = x1
POL(c9(x1, x2)) = x1 + x2
POL(chk(x1)) = [4]
POL(f(x1)) = [1]
POL(mark(x1)) = 0
POL(mat(x1, x2)) = x2
POL(no(x1)) = x1
POL(y) = 0

### (28) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(c)) → active(c)
mat(f(z0), c) → no(c)
mat(f(z0), f(y)) → f(mat(z0, y))
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c2, c9, c9

### (29) 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.

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
We considered the (Usable) Rules:

f(no(z0)) → no(f(z0))
active(f(z0)) → mark(f(f(z0)))
mat(f(z0), c) → no(c)
f(active(z0)) → active(f(z0))
f(mark(z0)) → mark(f(z0))
chk(no(c)) → active(c)
mat(f(z0), f(y)) → f(mat(z0, y))
And the Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVE(x1)) = [2]x1
POL(CHK(x1)) = [4]x1
POL(F(x1)) = [2]x1
POL(TP(x1)) = [4] + [4]x1
POL(X) = 0
POL(active(x1)) = [2] + [3]x1
POL(c) = 0
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(c8(x1)) = x1
POL(c9(x1)) = x1
POL(c9(x1, x2)) = x1 + x2
POL(chk(x1)) = [2]
POL(f(x1)) = [1] + [2]x1
POL(mark(x1)) = [2] + x1
POL(mat(x1, x2)) = [2]x2
POL(no(x1)) = x1
POL(y) = 0

### (30) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(active(z0)) → active(f(z0))
f(no(z0)) → no(f(z0))
f(mark(z0)) → mark(f(z0))
active(f(z0)) → mark(f(f(z0)))
chk(no(c)) → active(c)
mat(f(z0), c) → no(c)
mat(f(z0), f(y)) → f(mat(z0, y))
Tuples:

F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(no(z0)) → c7(F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
S tuples:none
K tuples:

F(no(z0)) → c7(F(z0))
CHK(no(f(f(y)))) → c2(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(f(y))) → c9(CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), f(y))))
TP(mark(c)) → c9(TP(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
CHK(no(f(c))) → c2(F(chk(no(c))), CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), c)))
F(active(z0)) → c6(ACTIVE(f(z0)), F(z0))
F(mark(z0)) → c8(F(z0))
ACTIVE(f(z0)) → c1(F(z0))
Defined Rule Symbols:

f, active, chk, mat

Defined Pair Symbols:

F, ACTIVE, CHK, TP

Compound Symbols:

c6, c7, c8, c1, c2, c2, c9, c9

### (31) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty