MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 60 seconds)' failed due to the following reason: Computation stopped due to timeout after 60.0 seconds. 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)' failed due to the following reason: The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: none TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [f](x1, x2, x3, x4, x5) = [1] x5 + [7] [s](x1) = [1] x1 + [7] [0] = [7] The following symbols are considered usable {f} The order satisfies the following ordering constraints: [f(s(x1), x2, x3, x4, x5)] = [1] x5 + [7] >= [1] x5 + [7] = [f(x1, x2, x3, x4, x5)] [f(0(), s(x2), x3, x4, x5)] = [1] x5 + [7] >= [1] x5 + [7] = [f(x2, x2, x3, x4, x5)] [f(0(), 0(), s(x3), x4, x5)] = [1] x5 + [7] >= [1] x5 + [7] = [f(x3, x3, x3, x4, x5)] [f(0(), 0(), 0(), s(x4), x5)] = [1] x5 + [7] >= [1] x5 + [7] = [f(x4, x4, x4, x4, x5)] [f(0(), 0(), 0(), 0(), s(x5))] = [1] x5 + [14] > [1] x5 + [7] = [f(x5, x5, x5, x5, x5)] [f(0(), 0(), 0(), 0(), 0())] = [14] > [7] = [0()] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) } Weak Trs: { f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Fastest' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: none TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1, x2, x3, x4, x5) = [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] [s](x1) = [1 2] x1 + [0] [0 1] [4] [0] = [0] [0] The following symbols are considered usable {f} The order satisfies the following ordering constraints: [f(s(x1), x2, x3, x4, x5)] = [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] >= [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] = [f(x1, x2, x3, x4, x5)] [f(0(), s(x2), x3, x4, x5)] = [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] >= [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] = [f(x2, x2, x3, x4, x5)] [f(0(), 0(), s(x3), x4, x5)] = [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] >= [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] = [f(x3, x3, x3, x4, x5)] [f(0(), 0(), 0(), s(x4), x5)] = [0 1] x4 + [4 0] x5 + [4] [0 0] [4 0] [4] > [0 1] x4 + [4 0] x5 + [0] [0 0] [4 0] [4] = [f(x4, x4, x4, x4, x5)] [f(0(), 0(), 0(), 0(), s(x5))] = [4 8] x5 + [0] [4 8] [4] >= [4 1] x5 + [0] [4 0] [4] = [f(x5, x5, x5, x5, x5)] [f(0(), 0(), 0(), 0(), 0())] = [0] [4] >= [0] [0] = [0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) } Weak Trs: { f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: We use the processor 'matrix interpretation of dimension 3' to orient following rules strictly. Trs: { f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^3)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: none TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [0 0 1] [0 1 0] [1 2 0] [0] [f](x1, x2, x3, x4, x5) = [0 0 0] x3 + [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0 0 0] [0] [1 4 0] [0] [s](x1) = [0 1 1] x1 + [0] [0 0 1] [1] [0] [0] = [0] [0] The following symbols are considered usable {f} The order satisfies the following ordering constraints: [f(s(x1), x2, x3, x4, x5)] = [0 0 1] [0 1 0] [1 2 0] [0] [0 0 0] x3 + [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0 0 0] [0] >= [0 0 1] [0 1 0] [1 2 0] [0] [0 0 0] x3 + [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0 0 0] [0] = [f(x1, x2, x3, x4, x5)] [f(0(), s(x2), x3, x4, x5)] = [0 0 1] [0 1 0] [1 2 0] [0] [0 0 0] x3 + [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0 0 0] [0] >= [0 0 1] [0 1 0] [1 2 0] [0] [0 0 0] x3 + [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0 0 0] [0] = [f(x2, x2, x3, x4, x5)] [f(0(), 0(), s(x3), x4, x5)] = [0 0 1] [0 1 0] [1 2 0] [1] [0 0 0] x3 + [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0 0 0] [0] > [0 0 1] [0 1 0] [1 2 0] [0] [0 0 0] x3 + [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0 0 0] [0] = [f(x3, x3, x3, x4, x5)] [f(0(), 0(), 0(), s(x4), x5)] = [0 1 1] [1 2 0] [0] [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0] >= [0 1 1] [1 2 0] [0] [0 0 0] x4 + [0 0 0] x5 + [4] [0 0 0] [0 0 0] [0] = [f(x4, x4, x4, x4, x5)] [f(0(), 0(), 0(), 0(), s(x5))] = [1 6 2] [0] [0 0 0] x5 + [4] [0 0 0] [0] >= [1 3 1] [0] [0 0 0] x5 + [4] [0 0 0] [0] = [f(x5, x5, x5, x5, x5)] [f(0(), 0(), 0(), 0(), 0())] = [0] [4] [0] >= [0] [0] [0] = [0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) } Weak Trs: { f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: We use the processor 'matrix interpretation of dimension 4' to orient following rules strictly. Trs: { f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^4)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: none TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [1 0 0 0] [0 0 0 1] [0 1 0 1] [1 1 1 1] [1] [f](x1, x2, x3, x4, x5) = [0 0 0 0] x2 + [0 0 0 0] x3 + [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [1 0 0 0] [1] [s](x1) = [0 1 0 1] x1 + [1] [0 1 1 1] [0] [1 0 0 1] [0] [0] [0] = [0] [0] [0] The following symbols are considered usable {f} The order satisfies the following ordering constraints: [f(s(x1), x2, x3, x4, x5)] = [1 0 0 0] [0 0 0 1] [0 1 0 1] [1 1 1 1] [1] [0 0 0 0] x2 + [0 0 0 0] x3 + [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] >= [1 0 0 0] [0 0 0 1] [0 1 0 1] [1 1 1 1] [1] [0 0 0 0] x2 + [0 0 0 0] x3 + [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] = [f(x1, x2, x3, x4, x5)] [f(0(), s(x2), x3, x4, x5)] = [1 0 0 0] [0 0 0 1] [0 1 0 1] [1 1 1 1] [2] [0 0 0 0] x2 + [0 0 0 0] x3 + [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] > [1 0 0 0] [0 0 0 1] [0 1 0 1] [1 1 1 1] [1] [0 0 0 0] x2 + [0 0 0 0] x3 + [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] = [f(x2, x2, x3, x4, x5)] [f(0(), 0(), s(x3), x4, x5)] = [1 0 0 1] [0 1 0 1] [1 1 1 1] [1] [0 0 0 0] x3 + [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] >= [1 0 0 1] [0 1 0 1] [1 1 1 1] [1] [0 0 0 0] x3 + [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0 0 0 0] [0] = [f(x3, x3, x3, x4, x5)] [f(0(), 0(), 0(), s(x4), x5)] = [1 1 0 2] [1 1 1 1] [2] [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0] > [1 1 0 2] [1 1 1 1] [1] [0 0 0 0] x4 + [0 0 0 0] x5 + [0] [0 0 0 0] [0 0 0 0] [0] [0 0 0 0] [0 0 0 0] [0] = [f(x4, x4, x4, x4, x5)] [f(0(), 0(), 0(), 0(), s(x5))] = [2 2 1 3] [3] [0 0 0 0] x5 + [0] [0 0 0 0] [0] [0 0 0 0] [0] > [2 2 1 3] [1] [0 0 0 0] x5 + [0] [0 0 0 0] [0] [0 0 0 0] [0] = [f(x5, x5, x5, x5, x5)] [f(0(), 0(), 0(), 0(), 0())] = [1] [0] [0] [0] > [0] [0] [0] [0] = [0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) } Weak Trs: { f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: We use the processor 'polynomial interpretation' to orient following rules strictly. Trs: { f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are considered usable: none TcT has computed the following constructor-restricted polynomial interpretation. [f](x1, x2, x3, x4, x5) = x4 + x5^2 [s](x1) = 1 + x1 [0]() = 0 The following symbols are considered usable {f} This order satisfies the following ordering constraints. [f(s(x1), x2, x3, x4, x5)] = x4 + x5^2 >= x4 + x5^2 = [f(x1, x2, x3, x4, x5)] [f(0(), s(x2), x3, x4, x5)] = x4 + x5^2 >= x4 + x5^2 = [f(x2, x2, x3, x4, x5)] [f(0(), 0(), s(x3), x4, x5)] = x4 + x5^2 >= x4 + x5^2 = [f(x3, x3, x3, x4, x5)] [f(0(), 0(), 0(), s(x4), x5)] = 1 + x4 + x5^2 > x4 + x5^2 = [f(x4, x4, x4, x4, x5)] [f(0(), 0(), 0(), 0(), s(x5))] = 1 + 2*x5 + x5^2 > x5 + x5^2 = [f(x5, x5, x5, x5, x5)] [f(0(), 0(), 0(), 0(), 0())] = >= = [0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) } Weak Trs: { f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: We use the processor 'polynomial interpretation' to orient following rules strictly. Trs: { f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^3)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are considered usable: none TcT has computed the following constructor-restricted polynomial interpretation. [f](x1, x2, x3, x4, x5) = 2*x3 + x4 + x4*x5 + x4^2 + x5^3 [s](x1) = 1 + x1 [0]() = 0 The following symbols are considered usable {f} This order satisfies the following ordering constraints. [f(s(x1), x2, x3, x4, x5)] = 2*x3 + x4 + x4*x5 + x4^2 + x5^3 >= 2*x3 + x4 + x4*x5 + x4^2 + x5^3 = [f(x1, x2, x3, x4, x5)] [f(0(), s(x2), x3, x4, x5)] = 2*x3 + x4 + x4*x5 + x4^2 + x5^3 >= 2*x3 + x4 + x4*x5 + x4^2 + x5^3 = [f(x2, x2, x3, x4, x5)] [f(0(), 0(), s(x3), x4, x5)] = 2 + 2*x3 + x4 + x4*x5 + x4^2 + x5^3 > 2*x3 + x4 + x4*x5 + x4^2 + x5^3 = [f(x3, x3, x3, x4, x5)] [f(0(), 0(), 0(), s(x4), x5)] = 2 + 3*x4 + x5 + x4*x5 + x4^2 + x5^3 > 3*x4 + x4*x5 + x4^2 + x5^3 = [f(x4, x4, x4, x4, x5)] [f(0(), 0(), 0(), 0(), s(x5))] = 1 + 3*x5 + 3*x5^2 + x5^3 > 3*x5 + 2*x5^2 + x5^3 = [f(x5, x5, x5, x5, x5)] [f(0(), 0(), 0(), 0(), 0())] = >= = [0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) } Weak Trs: { f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'bsearch-popstar (timeout of 60 seconds)' failed due to the following reason: The processor is inapplicable, reason: Processor only applicable for innermost runtime complexity analysis 2) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due to the following reason: The processor is inapplicable, reason: Processor only applicable for innermost runtime complexity analysis 3) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'Bounds with perSymbol-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 2) 'Bounds with minimal-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 3) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed due to the following reason: We add the following weak dependency pairs: Strict DPs: { f^#(s(x1), x2, x3, x4, x5) -> c_1(f^#(x1, x2, x3, x4, x5)) , f^#(0(), s(x2), x3, x4, x5) -> c_2(f^#(x2, x2, x3, x4, x5)) , f^#(0(), 0(), s(x3), x4, x5) -> c_3(f^#(x3, x3, x3, x4, x5)) , f^#(0(), 0(), 0(), s(x4), x5) -> c_4(f^#(x4, x4, x4, x4, x5)) , f^#(0(), 0(), 0(), 0(), s(x5)) -> c_5(f^#(x5, x5, x5, x5, x5)) , f^#(0(), 0(), 0(), 0(), 0()) -> c_6() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { f^#(s(x1), x2, x3, x4, x5) -> c_1(f^#(x1, x2, x3, x4, x5)) , f^#(0(), s(x2), x3, x4, x5) -> c_2(f^#(x2, x2, x3, x4, x5)) , f^#(0(), 0(), s(x3), x4, x5) -> c_3(f^#(x3, x3, x3, x4, x5)) , f^#(0(), 0(), 0(), s(x4), x5) -> c_4(f^#(x4, x4, x4, x4, x5)) , f^#(0(), 0(), 0(), 0(), s(x5)) -> c_5(f^#(x5, x5, x5, x5, x5)) , f^#(0(), 0(), 0(), 0(), 0()) -> c_6() } Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {6} by applications of Pre({6}) = {1,2,3,4,5}. Here rules are labeled as follows: DPs: { 1: f^#(s(x1), x2, x3, x4, x5) -> c_1(f^#(x1, x2, x3, x4, x5)) , 2: f^#(0(), s(x2), x3, x4, x5) -> c_2(f^#(x2, x2, x3, x4, x5)) , 3: f^#(0(), 0(), s(x3), x4, x5) -> c_3(f^#(x3, x3, x3, x4, x5)) , 4: f^#(0(), 0(), 0(), s(x4), x5) -> c_4(f^#(x4, x4, x4, x4, x5)) , 5: f^#(0(), 0(), 0(), 0(), s(x5)) -> c_5(f^#(x5, x5, x5, x5, x5)) , 6: f^#(0(), 0(), 0(), 0(), 0()) -> c_6() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { f^#(s(x1), x2, x3, x4, x5) -> c_1(f^#(x1, x2, x3, x4, x5)) , f^#(0(), s(x2), x3, x4, x5) -> c_2(f^#(x2, x2, x3, x4, x5)) , f^#(0(), 0(), s(x3), x4, x5) -> c_3(f^#(x3, x3, x3, x4, x5)) , f^#(0(), 0(), 0(), s(x4), x5) -> c_4(f^#(x4, x4, x4, x4, x5)) , f^#(0(), 0(), 0(), 0(), s(x5)) -> c_5(f^#(x5, x5, x5, x5, x5)) } Strict Trs: { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5) , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5) , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5) , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5) , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5) , f(0(), 0(), 0(), 0(), 0()) -> 0() } Weak DPs: { f^#(0(), 0(), 0(), 0(), 0()) -> c_6() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..