The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

(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
append(add(N, X), Y) → add(N, append(X, 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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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
SPLIT(z0, nil) → c5
F_2(true, z0, z1, z2, z3, z4) → c8
LT(0, s(z0)) → c
F_2(false, z0, z1, z2, z3, z4) → c9
LT(s(z0), 0) → c1
APPEND(nil, z0) → c3

(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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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(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:

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))
split(z0, nil) → pair(nil, nil)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)

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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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(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:

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))
split(z0, nil) → pair(nil, nil)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)

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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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(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:

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))
split(z0, nil) → pair(nil, nil)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)

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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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(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:

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))
split(z0, nil) → pair(nil, nil)
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)

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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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(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:

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))
split(z0, nil) → pair(nil, nil)
f_3(pair(z0, z1), z2, z3) → append(qsort(z0), add(z3, qsort(z1)))
qsort(nil) → nil
f_2(false, z0, z1, z2, z3, z4) → pair(add(z1, z3), z4)
append(nil, z0) → z0
split(z0, add(z1, z2)) → f_1(split(z0, z2), z0, z1, z2)
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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
append(add(z0, z1), z2) → add(z0, append(z1, z2))
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

(20) BOUNDS(1, 1)