(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^1).
The TRS R consists of the following rules:
g(S(x), y) → g(x, S(y))
f(y, S(x)) → f(S(y), x)
g(0, x2) → x2
f(x1, 0) → g(x1, 0)
Rewrite Strategy: INNERMOST
(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 2.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
S0(0) → 0
00() → 0
g0(0, 0) → 1
f0(0, 0) → 2
S1(0) → 3
g1(0, 3) → 1
S1(0) → 4
f1(4, 0) → 2
01() → 5
g1(0, 5) → 2
S1(3) → 3
S1(5) → 3
g1(0, 3) → 2
S1(4) → 4
g1(4, 5) → 2
S2(5) → 6
g2(0, 6) → 2
g2(4, 6) → 2
S1(6) → 3
S2(6) → 6
0 → 1
3 → 1
3 → 2
5 → 2
6 → 2
(2) BOUNDS(1, n^1)
(3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(S(z0), z1) → g(z0, S(z1))
g(0, z0) → z0
f(z0, S(z1)) → f(S(z0), z1)
f(z0, 0) → g(z0, 0)
Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
G(0, z0) → c1
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
G(0, z0) → c1
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c1, c2, c3
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
G(0, z0) → c1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(S(z0), z1) → g(z0, S(z1))
g(0, z0) → z0
f(z0, S(z1)) → f(S(z0), z1)
f(z0, 0) → g(z0, 0)
Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c2, c3
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
g(S(z0), z1) → g(z0, S(z1))
g(0, z0) → z0
f(z0, S(z1)) → f(S(z0), z1)
f(z0, 0) → g(z0, 0)
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
G, F
Compound Symbols:
c, c2, c3
(9) 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(z0, 0) → c3(G(z0, 0))
We considered the (Usable) Rules:none
And the Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(F(x1, x2)) = [1] + x1 + x2
POL(G(x1, x2)) = 0
POL(S(x1)) = x1
POL(c(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
K tuples:
F(z0, 0) → c3(G(z0, 0))
Defined Rule Symbols:none
Defined Pair Symbols:
G, F
Compound Symbols:
c, c2, c3
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
G(S(z0), z1) → c(G(z0, S(z1)))
We considered the (Usable) Rules:none
And the Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(F(x1, x2)) = [3] + [2]x1 + [2]x2
POL(G(x1, x2)) = x1
POL(S(x1)) = [2] + x1
POL(c(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:
F(z0, S(z1)) → c2(F(S(z0), z1))
K tuples:
F(z0, 0) → c3(G(z0, 0))
G(S(z0), z1) → c(G(z0, S(z1)))
Defined Rule Symbols:none
Defined Pair Symbols:
G, F
Compound Symbols:
c, c2, c3
(13) 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(z0, S(z1)) → c2(F(S(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(F(x1, x2)) = x2
POL(G(x1, x2)) = 0
POL(S(x1)) = [1] + x1
POL(c(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:none
K tuples:
F(z0, 0) → c3(G(z0, 0))
G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
Defined Rule Symbols:none
Defined Pair Symbols:
G, F
Compound Symbols:
c, c2, c3
(15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(16) BOUNDS(1, 1)