(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, 1).
The TRS R consists of the following rules:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(0)) → g(d(1))
g(c(1)) → g(d(0))
Rewrite Strategy: INNERMOST
(1) DependencyGraphProof (BOTH BOUNDS(ID, ID) transformation)
The following rules are not reachable from basic terms in the dependency graph and can be removed:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
(2) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, 1).
The TRS R consists of the following rules:
g(c(0)) → g(d(1))
g(d(x)) → x
g(c(x)) → x
g(c(1)) → g(d(0))
Rewrite Strategy: INNERMOST
(3) CpxTrsMatchBoundsProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2.
The certificate found is represented by the following graph.
Start state: 1
Accept states: [2]
Transitions:
1→2[g_1|0, 0|1, c_1|1, d_1|1, 1|1, 1|2, 0|2]
1→3[g_1|1]
1→5[g_1|1]
2→2[c_1|0, 0|0, d_1|0, 1|0]
3→4[d_1|1]
4→2[1|1]
5→6[d_1|1]
6→2[0|1]
(4) BOUNDS(1, n^1)
(5) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(c(0)) → g(d(1))
g(d(z0)) → z0
g(c(z0)) → z0
g(c(1)) → g(d(0))
Tuples:
G(c(0)) → c1(G(d(1)))
G(d(z0)) → c2
G(c(z0)) → c3
G(c(1)) → c4(G(d(0)))
S tuples:
G(c(0)) → c1(G(d(1)))
G(d(z0)) → c2
G(c(z0)) → c3
G(c(1)) → c4(G(d(0)))
K tuples:none
Defined Rule Symbols:
g
Defined Pair Symbols:
G
Compound Symbols:
c1, c2, c3, c4
(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing nodes:
G(c(0)) → c1(G(d(1)))
G(c(1)) → c4(G(d(0)))
G(c(z0)) → c3
G(d(z0)) → c2
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(c(0)) → g(d(1))
g(d(z0)) → z0
g(c(z0)) → z0
g(c(1)) → g(d(0))
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:
g
Defined Pair Symbols:none
Compound Symbols:none
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)