WORST_CASE(?,O(n^2))
proof of input_QiBFW5OJx2.trs
# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished


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

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
(3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
(4) CdtProblem
(5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
(6) CdtProblem
(7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 159 ms]
(8) CdtProblem
(9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 31 ms]
(10) CdtProblem
(11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 35 ms]
(12) CdtProblem
(13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 24 ms]
(14) CdtProblem
(15) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 2907 ms]
(16) CdtProblem
(17) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 2898 ms]
(18) CdtProblem
(19) SIsEmptyProof [BOTH BOUNDS(ID, ID), 0 ms]
(20) BOUNDS(1, 1)


----------------------------------------

(0)
Obligation:
The Runtime Complexity (parallel-innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

   f(S(x'), S(x)) -> h(g(x', S(x)), f(S(S(x')), x))
   h(0, S(x)) -> h(0, x)
   h(0, 0) -> 0
   g(S(x), S(x')) -> h(f(S(x), S(x')), g(x, S(S(x'))))
   g(S(x), 0) -> 0
   f(S(x), 0) -> 0
   h(S(x), x2) -> h(x, x2)
   g(0, x2) -> 0
   f(0, x2) -> 0

S is empty.
Rewrite Strategy: PARALLEL_INNERMOST
----------------------------------------

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
   f(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(S(z0), 0) -> 0
   g(0, z0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   F(S(z0), 0) -> c2
   F(0, z0) -> c3
   H(0, S(z0)) -> c4(H(0, z0))
   H(0, 0) -> c5
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
   G(S(z0), 0) -> c9
   G(0, z0) -> c10
S tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   F(S(z0), 0) -> c2
   F(0, z0) -> c3
   H(0, S(z0)) -> c4(H(0, z0))
   H(0, 0) -> c5
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
   G(S(z0), 0) -> c9
   G(0, z0) -> c10
K tuples:none
Defined Rule Symbols:   f_2, h_2, g_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c2, c3, c4_1, c5, c6_1, c7_2, c8_2, c9, c10


----------------------------------------

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 5 trailing nodes:
   F(0, z0) -> c3
   G(0, z0) -> c10
   F(S(z0), 0) -> c2
   G(S(z0), 0) -> c9
   H(0, 0) -> c5

----------------------------------------

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
   f(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(S(z0), 0) -> 0
   g(0, z0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
K tuples:none
Defined Rule Symbols:   f_2, h_2, g_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   f(0, z0) -> 0
   g(S(z0), 0) -> 0

----------------------------------------

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
K tuples:none
Defined Rule Symbols:   g_2, h_2, f_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
We considered the (Usable) Rules:none
And the Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = [1] + x_1 + x_2
   POL(G(x_1, x_2)) = [1] + x_1 + x_2
   POL(H(x_1, x_2)) = 0
   POL(S(x_1)) = [1] + x_1
   POL(c(x_1, x_2)) = x_1 + x_2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c4(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(g(x_1, x_2)) = 0
   POL(h(x_1, x_2)) = [1]

----------------------------------------

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
K tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
Defined Rule Symbols:   g_2, h_2, f_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
We considered the (Usable) Rules:none
And the Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = x_1 + x_2
   POL(G(x_1, x_2)) = [1] + x_1 + x_2
   POL(H(x_1, x_2)) = 0
   POL(S(x_1)) = [1] + x_1
   POL(c(x_1, x_2)) = x_1 + x_2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c4(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(g(x_1, x_2)) = 0
   POL(h(x_1, x_2)) = [1]

----------------------------------------

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
K tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
Defined Rule Symbols:   g_2, h_2, f_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   H(0, S(z0)) -> c4(H(0, z0))
We considered the (Usable) Rules:
   f(S(z0), 0) -> 0
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
And the Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = 0
   POL(G(x_1, x_2)) = 0
   POL(H(x_1, x_2)) = x_2
   POL(S(x_1)) = [1] + x_1
   POL(c(x_1, x_2)) = x_1 + x_2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c4(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(g(x_1, x_2)) = 0
   POL(h(x_1, x_2)) = 0

----------------------------------------

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
K tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   H(0, S(z0)) -> c4(H(0, z0))
Defined Rule Symbols:   g_2, h_2, f_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   H(S(z0), z1) -> c6(H(z0, z1))
We considered the (Usable) Rules:
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
And the Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = 0
   POL(G(x_1, x_2)) = 0
   POL(H(x_1, x_2)) = x_1
   POL(S(x_1)) = [1] + x_1
   POL(c(x_1, x_2)) = x_1 + x_2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c4(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(g(x_1, x_2)) = 0
   POL(h(x_1, x_2)) = 0

----------------------------------------

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
K tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
Defined Rule Symbols:   g_2, h_2, f_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(15) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
We considered the (Usable) Rules:none
And the Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = x_2 + x_2^2 + [2]x_1*x_2 + x_1^2
   POL(G(x_1, x_2)) = x_1 + x_2 + x_2^2 + [2]x_1*x_2 + x_1^2
   POL(H(x_1, x_2)) = 0
   POL(S(x_1)) = [2] + x_1
   POL(c(x_1, x_2)) = x_1 + x_2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c4(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(g(x_1, x_2)) = 0
   POL(h(x_1, x_2)) = 0

----------------------------------------

(16)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
K tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
Defined Rule Symbols:   g_2, h_2, f_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
We considered the (Usable) Rules:none
And the Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(F(x_1, x_2)) = x_2^2 + [2]x_1*x_2 + x_1^2
   POL(G(x_1, x_2)) = x_1 + x_2^2 + [2]x_1*x_2 + x_1^2
   POL(H(x_1, x_2)) = 0
   POL(S(x_1)) = [2] + x_1
   POL(c(x_1, x_2)) = x_1 + x_2
   POL(c1(x_1, x_2)) = x_1 + x_2
   POL(c4(x_1)) = x_1
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2)) = x_1 + x_2
   POL(c8(x_1, x_2)) = x_1 + x_2
   POL(f(x_1, x_2)) = 0
   POL(g(x_1, x_2)) = 0
   POL(h(x_1, x_2)) = 0

----------------------------------------

(18)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   g(S(z0), S(z1)) -> h(f(S(z0), S(z1)), g(z0, S(S(z1))))
   g(0, z0) -> 0
   h(0, S(z0)) -> h(0, z0)
   h(0, 0) -> 0
   h(S(z0), z1) -> h(z0, z1)
   f(S(z0), S(z1)) -> h(g(z0, S(z1)), f(S(S(z0)), z1))
   f(S(z0), 0) -> 0
Tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
S tuples:none
K tuples:
   F(S(z0), S(z1)) -> c(H(g(z0, S(z1)), f(S(S(z0)), z1)), G(z0, S(z1)))
   G(S(z0), S(z1)) -> c7(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), F(S(z0), S(z1)))
   H(0, S(z0)) -> c4(H(0, z0))
   H(S(z0), z1) -> c6(H(z0, z1))
   F(S(z0), S(z1)) -> c1(H(g(z0, S(z1)), f(S(S(z0)), z1)), F(S(S(z0)), z1))
   G(S(z0), S(z1)) -> c8(H(f(S(z0), S(z1)), g(z0, S(S(z1)))), G(z0, S(S(z1))))
Defined Rule Symbols:   g_2, h_2, f_2

Defined Pair Symbols:   F_2, H_2, G_2

Compound Symbols:   c_2, c1_2, c4_1, c6_1, c7_2, c8_2


----------------------------------------

(19) SIsEmptyProof (BOTH BOUNDS(ID, ID))
The set S is empty
----------------------------------------

(20)
BOUNDS(1, 1)