(0) Obligation:
The Runtime Complexity (full) of the given
CpxTRS could be proven to be
BOUNDS(1, 1).
The TRS R consists of the following rules:
f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, z) → z
Rewrite Strategy: FULL
(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)
The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
(2) Obligation:
The Runtime Complexity (full) of the given
CpxTRS could be proven to be
BOUNDS(1, 1).
The TRS R consists of the following rules:
b(y, z) → z
f(c(a, z, x)) → b(a, z)
Rewrite Strategy: FULL
(3) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)
Converted rc-obligation to irc-obligation.
As the TRS does not nest defined symbols, we have rc = irc.
(4) 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:
b(y, z) → z
f(c(a, z, x)) → b(a, z)
Rewrite Strategy: INNERMOST
(5) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
b(z0, z1) → z1
f(c(a, z0, z1)) → b(a, z0)
Tuples:
B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
S tuples:
B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
K tuples:none
Defined Rule Symbols:
b, f
Defined Pair Symbols:
B, F
Compound Symbols:
c1, c2
(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
B(z0, z1) → c1
F(c(a, z0, z1)) → c2(B(a, z0))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
b(z0, z1) → z1
f(c(a, z0, z1)) → b(a, z0)
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:
b, f
Defined Pair Symbols:none
Compound Symbols:none
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)