We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) , quot(0(), s(y)) -> 0() , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , plus(minus(x, s(0())), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0()))) , plus(0(), y) -> y , plus(s(x), y) -> s(plus(x, y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(0(), y) -> c_6() , plus^#(s(x), y) -> c_7(plus^#(x, y)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(0(), y) -> c_6() , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Strict Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) , quot(0(), s(y)) -> 0() , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , plus(minus(x, s(0())), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0()))) , plus(0(), y) -> y , plus(s(x), y) -> s(plus(x, y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(0(), y) -> c_6() , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Strict Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_2) = {1}, Uargs(quot^#) = {1}, Uargs(c_4) = {1}, Uargs(c_7) = {1} TcT has computed the following constructor-restricted matrix interpretation. [minus](x1, x2) = [1 1] x1 + [2] [2 2] [0] [0] = [0] [0] [s](x1) = [1 1] x1 + [0] [0 0] [2] [minus^#](x1, x2) = [0] [0] [c_1] = [0] [0] [c_2](x1) = [1 0] x1 + [0] [0 1] [0] [quot^#](x1, x2) = [1 0] x1 + [0] [0 0] [0] [c_3] = [0] [0] [c_4](x1) = [1 0] x1 + [0] [0 1] [0] [plus^#](x1, x2) = [0] [0] [c_5](x1) = [0] [0] [c_6] = [0] [0] [c_7](x1) = [1 0] x1 + [0] [0 1] [0] The order satisfies the following ordering constraints: [minus(x, 0())] = [1 1] x + [2] [2 2] [0] > [1 0] x + [0] [0 1] [0] = [x] [minus(s(x), s(y))] = [1 1] x + [4] [2 2] [4] > [1 1] x + [2] [2 2] [0] = [minus(x, y)] [minus^#(x, 0())] = [0] [0] >= [0] [0] = [c_1()] [minus^#(s(x), s(y))] = [0] [0] >= [0] [0] = [c_2(minus^#(x, y))] [quot^#(0(), s(y))] = [0] [0] >= [0] [0] = [c_3()] [quot^#(s(x), s(y))] = [1 1] x + [0] [0 0] [0] ? [1 1] x + [2] [0 0] [0] = [c_4(quot^#(minus(x, y), s(y)))] [plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [0] [0] >= [0] [0] = [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))] [plus^#(0(), y)] = [0] [0] >= [0] [0] = [c_6()] [plus^#(s(x), y)] = [0] [0] >= [0] [0] = [c_7(plus^#(x, y))] 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 YES(O(1),O(n^1)). Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(0(), y) -> c_6() , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1,3,6} by applications of Pre({1,3,6}) = {2,4,5,7}. Here rules are labeled as follows: DPs: { 1: minus^#(x, 0()) -> c_1() , 2: minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , 3: quot^#(0(), s(y)) -> c_3() , 4: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , 5: plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , 6: plus^#(0(), y) -> c_6() , 7: plus^#(s(x), y) -> c_7(plus^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Weak DPs: { minus^#(x, 0()) -> c_1() , quot^#(0(), s(y)) -> c_3() , plus^#(0(), y) -> c_6() } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { minus^#(x, 0()) -> c_1() , quot^#(0(), s(y)) -> c_3() , plus^#(0(), y) -> c_6() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 3: plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_4) = {1}, Uargs(c_7) = {1} TcT has computed the following constructor-restricted matrix interpretation. Note that the diagonal of the component-wise maxima of interpretation-entries (of constructors) contains no more than 0 non-zero entries. [minus](x1, x2) = [0] [0] = [0] [s](x1) = [0] [minus^#](x1, x2) = [0] [c_2](x1) = [4] x1 + [0] [quot^#](x1, x2) = [0] [c_4](x1) = [4] x1 + [0] [plus^#](x1, x2) = [1] x2 + [1] [c_5](x1) = [0] [c_7](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [minus(x, 0())] = [0] ? [1] x + [0] = [x] [minus(s(x), s(y))] = [0] >= [0] = [minus(x, y)] [minus^#(s(x), s(y))] = [0] >= [0] = [c_2(minus^#(x, y))] [quot^#(s(x), s(y))] = [0] >= [0] = [c_4(quot^#(minus(x, y), s(y)))] [plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [1] > [0] = [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))] [plus^#(s(x), y)] = [1] y + [1] >= [1] y + [1] = [c_7(plus^#(x, y))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Weak DPs: { plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , 4: plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_4) = {1}, Uargs(c_7) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [minus](x1, x2) = [0] [0] = [0] [s](x1) = [1] x1 + [4] [minus^#](x1, x2) = [2] x1 + [0] [c_2](x1) = [1] x1 + [1] [quot^#](x1, x2) = [0] [c_4](x1) = [1] x1 + [0] [plus^#](x1, x2) = [1] x2 + [4] [c_5](x1) = [0] [c_7](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [minus(x, 0())] = [0] ? [1] x + [0] = [x] [minus(s(x), s(y))] = [0] >= [0] = [minus(x, y)] [minus^#(s(x), s(y))] = [2] x + [8] > [2] x + [1] = [c_2(minus^#(x, y))] [quot^#(s(x), s(y))] = [0] >= [0] = [c_4(quot^#(minus(x, y), s(y)))] [plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [4] > [0] = [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))] [plus^#(s(x), y)] = [1] y + [4] >= [1] y + [4] = [c_7(plus^#(x, y))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Weak DPs: { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Weak DPs: { plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 2: plus^#(s(x), y) -> c_7(plus^#(x, y)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_4) = {1}, Uargs(c_7) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [minus](x1, x2) = [0] [0] = [0] [s](x1) = [1] x1 + [5] [minus^#](x1, x2) = [0] [c_2](x1) = [0] [quot^#](x1, x2) = [2] [c_4](x1) = [1] x1 + [0] [plus^#](x1, x2) = [1] x1 + [1] x2 + [0] [c_5](x1) = [0] [c_7](x1) = [1] x1 + [0] The order satisfies the following ordering constraints: [minus(x, 0())] = [0] ? [1] x + [0] = [x] [minus(s(x), s(y))] = [0] >= [0] = [minus(x, y)] [quot^#(s(x), s(y))] = [2] >= [2] = [c_4(quot^#(minus(x, y), s(y)))] [plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [0] >= [0] = [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))] [plus^#(s(x), y)] = [1] x + [1] y + [5] > [1] x + [1] y + [0] = [c_7(plus^#(x, y))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Weak DPs: { plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(s(x), y) -> c_7(plus^#(x, y)) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { plus^#(minus(x, s(0())), minus(y, s(s(z)))) -> c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) , plus^#(s(x), y) -> c_7(plus^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'Small Polynomial Path Order (PS,1-bounded)' to orient following rules strictly. DPs: { 1: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Trs: { minus(s(x), s(y)) -> minus(x, y) } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(minus) = {1, 2}, safe(0) = {}, safe(s) = {1}, safe(minus^#) = {}, safe(c_2) = {}, safe(quot^#) = {2}, safe(c_4) = {}, safe(plus^#) = {}, safe(c_5) = {}, safe(c_7) = {} and precedence empty . Following symbols are considered recursive: {quot^#} The recursion depth is 1. Further, following argument filtering is employed: pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(minus^#) = [], pi(c_2) = [], pi(quot^#) = [1, 2], pi(c_4) = [1], pi(plus^#) = [], pi(c_5) = [], pi(c_7) = [] Usable defined function symbols are a subset of: {minus, minus^#, quot^#, plus^#} For your convenience, here are the satisfied ordering constraints: pi(quot^#(s(x), s(y))) = quot^#(s(; x); s(; y)) > c_4(quot^#(x; s(; y));) = pi(c_4(quot^#(minus(x, y), s(y)))) pi(minus(x, 0())) = x >= x = pi(x) pi(minus(s(x), s(y))) = s(; x) > x = pi(minus(x, y)) The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))