### (0) Obligation:

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

lt(0, s(X)) → true
lt(s(X), 0) → false
lt(s(X), s(Y)) → lt(X, Y)
append(nil, Y) → Y
split(N, nil) → pair(nil, nil)
split(N, add(M, Y)) → f_1(split(N, Y), N, M, Y)
f_1(pair(X, Z), N, M, Y) → f_2(lt(N, M), N, M, Y, X, Z)
f_2(true, N, M, Y, X, Z) → pair(X, add(M, Z))
f_2(false, N, M, Y, X, Z) → pair(add(M, X), Z)
qsort(nil) → nil
qsort(add(N, X)) → f_3(split(N, X), N, X)
f_3(pair(Y, Z), N, X) → append(qsort(Y), add(X, qsort(Z)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(0, s(z0)) → c
LT(s(z0), 0) → c1
LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(nil, z0) → c3
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, nil) → c5
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
F_1(pair(z0, z1), z2, z3, z4) → c7(F_2(lt(z2, z3), z2, z3, z4, z0, z1), LT(z2, z3))
F_2(true, z0, z1, z2, z3, z4) → c8
F_2(false, z0, z1, z2, z3, z4) → c9
QSORT(nil) → c10
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
S tuples:

LT(0, s(z0)) → c
LT(s(z0), 0) → c1
LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(nil, z0) → c3
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, nil) → c5
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
F_1(pair(z0, z1), z2, z3, z4) → c7(F_2(lt(z2, z3), z2, z3, z4, z0, z1), LT(z2, z3))
F_2(true, z0, z1, z2, z3, z4) → c8
F_2(false, z0, z1, z2, z3, z4) → c9
QSORT(nil) → c10
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
K tuples:none
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, F_1, F_2, QSORT, F_3

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12

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

Removed 7 trailing nodes:

QSORT(nil) → c10
LT(0, s(z0)) → c
LT(s(z0), 0) → c1
SPLIT(z0, nil) → c5
APPEND(nil, z0) → c3
F_2(false, z0, z1, z2, z3, z4) → c9
F_2(true, z0, z1, z2, z3, z4) → c8

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
F_1(pair(z0, z1), z2, z3, z4) → c7(F_2(lt(z2, z3), z2, z3, z4, z0, z1), LT(z2, z3))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
S tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
F_1(pair(z0, z1), z2, z3, z4) → c7(F_2(lt(z2, z3), z2, z3, z4, z0, z1), LT(z2, z3))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
K tuples:none
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, F_1, QSORT, F_3

Compound Symbols:

c2, c4, c6, c7, c11, c12

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

Removed 1 trailing tuple parts

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
S tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
K tuples:none
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, QSORT, F_3, F_1

Compound Symbols:

c2, c4, c6, c11, c12, c7

### (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.

F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
We considered the (Usable) Rules:

split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
split(z0, nil) → pair(nil, nil)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
And the Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [2]
POL(APPEND(x1, x2)) = [1]
POL(F_1(x1, x2, x3, x4)) = 0
POL(F_3(x1, x2, x3)) = [2] + x1
POL(LT(x1, x2)) = 0
POL(QSORT(x1)) = x1
POL(SPLIT(x1, x2)) = 0
POL(add(x1, x2)) = [2] + x2
POL(append(x1, x2)) = [2]
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(f_1(x1, x2, x3, x4)) = [2] + x1
POL(f_2(x1, x2, x3, x4, x5, x6)) = [2] + x5 + x6
POL(f_3(x1, x2, x3)) = [3] + x2
POL(false) = 0
POL(lt(x1, x2)) = 0
POL(nil) = 0
POL(pair(x1, x2)) = x1 + x2
POL(qsort(x1)) = 0
POL(s(x1)) = [3]
POL(split(x1, x2)) = x2
POL(true) = 0

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
S tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
K tuples:

F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, QSORT, F_3, F_1

Compound Symbols:

c2, c4, c6, c11, c12, c7

### (9) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
S tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
K tuples:

F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, QSORT, F_3, F_1

Compound Symbols:

c2, c4, c6, c11, c12, c7

### (11) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
We considered the (Usable) Rules:

split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
split(z0, nil) → pair(nil, nil)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
And the Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APPEND(x1, x2)) = 0
POL(F_1(x1, x2, x3, x4)) = [1]
POL(F_3(x1, x2, x3)) = x12
POL(LT(x1, x2)) = 0
POL(QSORT(x1)) = x12
POL(SPLIT(x1, x2)) = x2
POL(add(x1, x2)) = [1] + x2
POL(append(x1, x2)) = 0
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(f_1(x1, x2, x3, x4)) = [1] + x1
POL(f_2(x1, x2, x3, x4, x5, x6)) = [1] + x5 + x6
POL(f_3(x1, x2, x3)) = 0
POL(false) = 0
POL(lt(x1, x2)) = 0
POL(nil) = 0
POL(pair(x1, x2)) = x1 + x2
POL(qsort(x1)) = 0
POL(s(x1)) = 0
POL(split(x1, x2)) = x2
POL(true) = 0

### (12) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
S tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
K tuples:

F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, QSORT, F_3, F_1

Compound Symbols:

c2, c4, c6, c11, c12, c7

### (13) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
We considered the (Usable) Rules:

split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
split(z0, nil) → pair(nil, nil)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
And the Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APPEND(x1, x2)) = [1]
POL(F_1(x1, x2, x3, x4)) = [1]
POL(F_3(x1, x2, x3)) = [1] + [2]x3 + [2]x12
POL(LT(x1, x2)) = [1]
POL(QSORT(x1)) = [2]x12
POL(SPLIT(x1, x2)) = [1] + [2]x2
POL(add(x1, x2)) = [1] + x2
POL(append(x1, x2)) = 0
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(f_1(x1, x2, x3, x4)) = [1] + x1
POL(f_2(x1, x2, x3, x4, x5, x6)) = [1] + x5 + x6
POL(f_3(x1, x2, x3)) = 0
POL(false) = 0
POL(lt(x1, x2)) = 0
POL(nil) = 0
POL(pair(x1, x2)) = x1 + x2
POL(qsort(x1)) = 0
POL(s(x1)) = 0
POL(split(x1, x2)) = x2
POL(true) = 0

### (14) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
S tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
K tuples:

F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, QSORT, F_3, F_1

Compound Symbols:

c2, c4, c6, c11, c12, c7

### (15) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

LT(s(z0), s(z1)) → c2(LT(z0, z1))
We considered the (Usable) Rules:

split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
split(z0, nil) → pair(nil, nil)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
And the Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APPEND(x1, x2)) = 0
POL(F_1(x1, x2, x3, x4)) = [1] + x3
POL(F_3(x1, x2, x3)) = [1] + x2 + x3 + x12
POL(LT(x1, x2)) = x2
POL(QSORT(x1)) = x12
POL(SPLIT(x1, x2)) = x1 + x2 + x12
POL(add(x1, x2)) = [1] + x1 + x2
POL(append(x1, x2)) = 0
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(f_1(x1, x2, x3, x4)) = [1] + x1 + x3
POL(f_2(x1, x2, x3, x4, x5, x6)) = [1] + x3 + x5 + x6
POL(f_3(x1, x2, x3)) = 0
POL(false) = 0
POL(lt(x1, x2)) = 0
POL(nil) = 0
POL(pair(x1, x2)) = x1 + x2
POL(qsort(x1)) = 0
POL(s(x1)) = [2] + x1
POL(split(x1, x2)) = x2
POL(true) = 0

### (16) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
S tuples:

APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
K tuples:

F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
LT(s(z0), s(z1)) → c2(LT(z0, z1))
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, QSORT, F_3, F_1

Compound Symbols:

c2, c4, c6, c11, c12, c7

### (17) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
We considered the (Usable) Rules:

append(nil, z0) → z0
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
qsort(nil) → nil
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
split(z0, nil) → pair(nil, nil)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
And the Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0
POL(APPEND(x1, x2)) = [1] + x1
POL(F_1(x1, x2, x3, x4)) = 0
POL(F_3(x1, x2, x3)) = [2] + [2]x1 + x12
POL(LT(x1, x2)) = 0
POL(QSORT(x1)) = x12
POL(SPLIT(x1, x2)) = [1]
POL(add(x1, x2)) = [2] + x2
POL(append(x1, x2)) = x1 + x2
POL(c11(x1, x2)) = x1 + x2
POL(c12(x1, x2, x3)) = x1 + x2 + x3
POL(c2(x1)) = x1
POL(c4(x1)) = x1
POL(c6(x1, x2)) = x1 + x2
POL(c7(x1)) = x1
POL(f_1(x1, x2, x3, x4)) = [2] + x1
POL(f_2(x1, x2, x3, x4, x5, x6)) = [2] + x5 + x6
POL(f_3(x1, x2, x3)) = [2] + [2]x1
POL(false) = 0
POL(lt(x1, x2)) = 0
POL(nil) = 0
POL(pair(x1, x2)) = x1 + x2
POL(qsort(x1)) = [2]x1
POL(s(x1)) = 0
POL(split(x1, x2)) = x2
POL(true) = 0

### (18) Obligation:

Complexity Dependency Tuples Problem
Rules:

lt(0, s(z0)) → true
lt(s(z0), 0) → false
lt(s(z0), s(z1)) → lt(z0, z1)
append(nil, z0) → z0
split(z0, nil) → pair(nil, nil)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
f_1(pair(z0, z1), z2, z3, z4) → f_2(lt(z2, z3), z2, z3, z4, z0, z1)
f_2(true, z0, z1, z2, z3, z4) → pair(z3, add(z1, z4))
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
qsort(nil) → nil
qsort(add(z0, z1)) → f_3(split(z0, z1), z0, z1)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
Tuples:

LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
S tuples:none
K tuples:

F_3(pair(z0, z1), z2, z3) → c12(APPEND(qsort(z0), add(z3, qsort(z1))), QSORT(z0), QSORT(z1))
QSORT(add(z0, z1)) → c11(F_3(split(z0, z1), z0, z1), SPLIT(z0, z1))
F_1(pair(z0, z1), z2, z3, z4) → c7(LT(z2, z3))
SPLIT(z0, add(z1, z2)) → c6(F_1(split(z0, z2), z0, z1, z2), SPLIT(z0, z2))
LT(s(z0), s(z1)) → c2(LT(z0, z1))
APPEND(add(z0, z1), z2) → c4(APPEND(z1, z2))
Defined Rule Symbols:

lt, append, split, f_1, f_2, qsort, f_3

Defined Pair Symbols:

LT, APPEND, SPLIT, QSORT, F_3, F_1

Compound Symbols:

c2, c4, c6, c11, c12, c7

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

The set S is empty