YES(O(1), O(n^1)) 3.13/1.25 YES(O(1), O(n^1)) 3.13/1.29 3.13/1.29 3.13/1.29 3.13/1.29 3.13/1.29 3.13/1.29 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 3.13/1.29 3.13/1.29 3.13/1.29
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3.13/1.29
3.13/1.29
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
3.13/1.29 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
3.13/1.29 
↳2 CdtProblem
3.13/1.29 
↳3 CdtUnreachableProof (⇔)
3.13/1.29 
↳4 CdtProblem
3.13/1.29 
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
3.13/1.29 
↳6 CdtProblem
3.13/1.29 
↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
3.13/1.29 
↳8 CdtProblem
3.13/1.29 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.13/1.29 
↳10 CdtProblem
3.13/1.29 
↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.13/1.29 
↳12 CdtProblem
3.13/1.29 
↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
3.13/1.29 
↳14 CdtProblem
3.13/1.29 
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID))
3.13/1.29 
↳16 BOUNDS(O(1), O(1))
3.13/1.29
3.13/1.29 3.13/1.29
3.13/1.29
3.13/1.29

(0) Obligation:

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

t(N) → cs(r(q(N)), nt(ns(N))) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(X)) → s(p(q(X), d(X))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(X)) → s(s(d(X))) 3.13/1.29
p(0, X) → X 3.13/1.29
p(X, 0) → X 3.13/1.29
p(s(X), s(Y)) → s(s(p(X, Y))) 3.13/1.29
f(0, X) → nil 3.13/1.29
f(s(X), cs(Y, Z)) → cs(Y, nf(X, a(Z))) 3.13/1.29
t(X) → nt(X) 3.13/1.29
s(X) → ns(X) 3.13/1.29
f(X1, X2) → nf(X1, X2) 3.13/1.29
a(nt(X)) → t(a(X)) 3.13/1.29
a(ns(X)) → s(a(X)) 3.13/1.29
a(nf(X1, X2)) → f(a(X1), a(X2)) 3.13/1.29
a(X) → X

Rewrite Strategy: INNERMOST
3.13/1.29
3.13/1.29

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

Converted CpxTRS to CDT
3.13/1.29
3.13/1.29

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(z0)) → s(p(q(z0), d(z0))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(z0)) → s(s(d(z0))) 3.13/1.29
p(0, z0) → z0 3.13/1.29
p(z0, 0) → z0 3.13/1.29
p(s(z0), s(z1)) → s(s(p(z0, z1))) 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2))) 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0
Tuples:

T(z0) → c(Q(z0)) 3.13/1.29
Q(s(z0)) → c3(S(p(q(z0), d(z0))), P(q(z0), d(z0)), Q(z0), D(z0)) 3.13/1.29
D(s(z0)) → c5(S(s(d(z0))), S(d(z0)), D(z0)) 3.13/1.29
P(s(z0), s(z1)) → c8(S(s(p(z0, z1))), S(p(z0, z1)), P(z0, z1)) 3.13/1.29
F(s(z0), cs(z1, z2)) → c10(A(z2)) 3.13/1.29
A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
A(ns(z0)) → c14(S(a(z0)), A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))
S tuples:

T(z0) → c(Q(z0)) 3.13/1.29
Q(s(z0)) → c3(S(p(q(z0), d(z0))), P(q(z0), d(z0)), Q(z0), D(z0)) 3.13/1.29
D(s(z0)) → c5(S(s(d(z0))), S(d(z0)), D(z0)) 3.13/1.29
P(s(z0), s(z1)) → c8(S(s(p(z0, z1))), S(p(z0, z1)), P(z0, z1)) 3.13/1.29
F(s(z0), cs(z1, z2)) → c10(A(z2)) 3.13/1.29
A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
A(ns(z0)) → c14(S(a(z0)), A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))
K tuples:none
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

T, Q, D, P, F, A

Compound Symbols:

c, c3, c5, c8, c10, c13, c14, c15

3.13/1.29
3.13/1.29

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

Q(s(z0)) → c3(S(p(q(z0), d(z0))), P(q(z0), d(z0)), Q(z0), D(z0)) 3.13/1.29
D(s(z0)) → c5(S(s(d(z0))), S(d(z0)), D(z0)) 3.13/1.29
P(s(z0), s(z1)) → c8(S(s(p(z0, z1))), S(p(z0, z1)), P(z0, z1)) 3.13/1.29
F(s(z0), cs(z1, z2)) → c10(A(z2))
3.13/1.29
3.13/1.29

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(z0)) → s(p(q(z0), d(z0))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(z0)) → s(s(d(z0))) 3.13/1.29
p(0, z0) → z0 3.13/1.29
p(z0, 0) → z0 3.13/1.29
p(s(z0), s(z1)) → s(s(p(z0, z1))) 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2))) 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0
Tuples:

T(z0) → c(Q(z0)) 3.13/1.29
A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
A(ns(z0)) → c14(S(a(z0)), A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))
S tuples:

T(z0) → c(Q(z0)) 3.13/1.29
A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
A(ns(z0)) → c14(S(a(z0)), A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(F(a(z0), a(z1)), A(z0), A(z1))
K tuples:none
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

T, A

Compound Symbols:

c, c13, c14, c15

3.13/1.29
3.13/1.29

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

Removed 3 trailing tuple parts
3.13/1.29
3.13/1.29

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(z0)) → s(p(q(z0), d(z0))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(z0)) → s(s(d(z0))) 3.13/1.29
p(0, z0) → z0 3.13/1.29
p(z0, 0) → z0 3.13/1.29
p(s(z0), s(z1)) → s(s(p(z0, z1))) 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2))) 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0
Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
S tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
K tuples:none
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A, T

Compound Symbols:

c13, c, c14, c15

3.13/1.29
3.13/1.29

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

Removed 1 trailing nodes:

T(z0) → c
3.13/1.29
3.13/1.29

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(z0)) → s(p(q(z0), d(z0))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(z0)) → s(s(d(z0))) 3.13/1.29
p(0, z0) → z0 3.13/1.29
p(z0, 0) → z0 3.13/1.29
p(s(z0), s(z1)) → s(s(p(z0, z1))) 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2))) 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0
Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
S tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
K tuples:none
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A, T

Compound Symbols:

c13, c, c14, c15

3.13/1.29
3.13/1.29

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

A(ns(z0)) → c14(A(z0))
We considered the (Usable) Rules:

a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0
And the Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.13/1.29

POL(0) = 0    3.13/1.29
POL(A(x1)) = x1    3.13/1.29
POL(T(x1)) = 0    3.13/1.29
POL(a(x1)) = 0    3.13/1.29
POL(c) = 0    3.13/1.29
POL(c13(x1, x2)) = x1 + x2    3.13/1.29
POL(c14(x1)) = x1    3.13/1.29
POL(c15(x1, x2)) = x1 + x2    3.13/1.29
POL(cs(x1, x2)) = [3] + x1    3.13/1.29
POL(f(x1, x2)) = [3]    3.13/1.29
POL(nf(x1, x2)) = x1 + x2    3.13/1.29
POL(nil) = [3]    3.13/1.29
POL(ns(x1)) = [1] + x1    3.13/1.29
POL(nt(x1)) = x1    3.13/1.29
POL(q(x1)) = [1]    3.13/1.29
POL(r(x1)) = [3] + x1    3.13/1.29
POL(s(x1)) = [3]    3.13/1.29
POL(t(x1)) = [3]   
3.13/1.29
3.13/1.29

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(z0)) → s(p(q(z0), d(z0))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(z0)) → s(s(d(z0))) 3.13/1.29
p(0, z0) → z0 3.13/1.29
p(z0, 0) → z0 3.13/1.29
p(s(z0), s(z1)) → s(s(p(z0, z1))) 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2))) 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0
Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
S tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
K tuples:

A(ns(z0)) → c14(A(z0))
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A, T

Compound Symbols:

c13, c, c14, c15

3.13/1.29
3.13/1.29

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

A(nf(z0, z1)) → c15(A(z0), A(z1))
We considered the (Usable) Rules:

a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0
And the Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.13/1.29

POL(0) = [2]    3.13/1.29
POL(A(x1)) = [5] + [4]x1    3.13/1.29
POL(T(x1)) = 0    3.13/1.29
POL(a(x1)) = 0    3.13/1.29
POL(c) = 0    3.13/1.29
POL(c13(x1, x2)) = x1 + x2    3.13/1.29
POL(c14(x1)) = x1    3.13/1.29
POL(c15(x1, x2)) = x1 + x2    3.13/1.29
POL(cs(x1, x2)) = [3] + x1    3.13/1.29
POL(f(x1, x2)) = [3]    3.13/1.29
POL(nf(x1, x2)) = [4] + x1 + x2    3.13/1.29
POL(nil) = [3]    3.13/1.29
POL(ns(x1)) = x1    3.13/1.29
POL(nt(x1)) = x1    3.13/1.29
POL(q(x1)) = [4]    3.13/1.29
POL(r(x1)) = [3] + x1    3.13/1.29
POL(s(x1)) = [3]    3.13/1.29
POL(t(x1)) = [3]   
3.13/1.29
3.13/1.29

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(z0)) → s(p(q(z0), d(z0))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(z0)) → s(s(d(z0))) 3.13/1.29
p(0, z0) → z0 3.13/1.29
p(z0, 0) → z0 3.13/1.29
p(s(z0), s(z1)) → s(s(p(z0, z1))) 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2))) 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0
Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
S tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c
K tuples:

A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A, T

Compound Symbols:

c13, c, c14, c15

3.13/1.29
3.13/1.29

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c
We considered the (Usable) Rules:

a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0
And the Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.13/1.29

POL(0) = [2]    3.13/1.29
POL(A(x1)) = [2]x1    3.13/1.29
POL(T(x1)) = [3]    3.13/1.29
POL(a(x1)) = [2]x1    3.13/1.29
POL(c) = 0    3.13/1.29
POL(c13(x1, x2)) = x1 + x2    3.13/1.29
POL(c14(x1)) = x1    3.13/1.29
POL(c15(x1, x2)) = x1 + x2    3.13/1.29
POL(cs(x1, x2)) = [3] + x1    3.13/1.29
POL(f(x1, x2)) = x1 + x2    3.13/1.29
POL(nf(x1, x2)) = x1 + x2    3.13/1.29
POL(nil) = [2]    3.13/1.29
POL(ns(x1)) = x1    3.13/1.29
POL(nt(x1)) = [2] + x1    3.13/1.29
POL(q(x1)) = [4]    3.13/1.29
POL(r(x1)) = [1]    3.13/1.29
POL(s(x1)) = x1    3.13/1.29
POL(t(x1)) = [4] + x1   
3.13/1.29
3.13/1.29

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

t(z0) → cs(r(q(z0)), nt(ns(z0))) 3.13/1.29
t(z0) → nt(z0) 3.13/1.29
q(0) → 0 3.13/1.29
q(s(z0)) → s(p(q(z0), d(z0))) 3.13/1.29
d(0) → 0 3.13/1.29
d(s(z0)) → s(s(d(z0))) 3.13/1.29
p(0, z0) → z0 3.13/1.29
p(z0, 0) → z0 3.13/1.29
p(s(z0), s(z1)) → s(s(p(z0, z1))) 3.13/1.29
f(0, z0) → nil 3.13/1.29
f(s(z0), cs(z1, z2)) → cs(z1, nf(z0, a(z2))) 3.13/1.29
f(z0, z1) → nf(z0, z1) 3.13/1.29
s(z0) → ns(z0) 3.13/1.29
a(nt(z0)) → t(a(z0)) 3.13/1.29
a(ns(z0)) → s(a(z0)) 3.13/1.29
a(nf(z0, z1)) → f(a(z0), a(z1)) 3.13/1.29
a(z0) → z0
Tuples:

A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c 3.13/1.29
A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1))
S tuples:none
K tuples:

A(ns(z0)) → c14(A(z0)) 3.13/1.29
A(nf(z0, z1)) → c15(A(z0), A(z1)) 3.13/1.29
A(nt(z0)) → c13(T(a(z0)), A(z0)) 3.13/1.29
T(z0) → c
Defined Rule Symbols:

t, q, d, p, f, s, a

Defined Pair Symbols:

A, T

Compound Symbols:

c13, c, c14, c15

3.13/1.29
3.13/1.29

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

The set S is empty
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3.13/1.29

(16) BOUNDS(O(1), O(1))

3.13/1.29
3.13/1.29
3.49/1.39 EOF