YES(O(1),O(n^2)) 166.20/60.07 YES(O(1),O(n^2)) 166.20/60.07 166.20/60.07 We are left with following problem, upon which TcT provides the 166.20/60.07 certificate YES(O(1),O(n^2)). 166.20/60.07 166.20/60.07 Strict Trs: 166.20/60.07 { or(x, x) -> x 166.20/60.07 , and(x, x) -> x 166.20/60.07 , not(or(x, y)) -> and(not(x), not(y)) 166.20/60.07 , not(and(x, y)) -> or(not(x), not(y)) 166.20/60.07 , not(not(x)) -> x } 166.20/60.07 Obligation: 166.20/60.07 derivational complexity 166.20/60.07 Answer: 166.20/60.07 YES(O(1),O(n^2)) 166.20/60.07 166.20/60.07 We use the processor 'matrix interpretation of dimension 1' to 166.20/60.07 orient following rules strictly. 166.20/60.07 166.20/60.07 Trs: 166.20/60.07 { or(x, x) -> x 166.20/60.07 , and(x, x) -> x } 166.20/60.07 166.20/60.07 The induced complexity on above rules (modulo remaining rules) is 166.20/60.07 YES(?,O(n^1)) . These rules are moved into the corresponding weak 166.20/60.07 component(s). 166.20/60.07 166.20/60.07 Sub-proof: 166.20/60.07 ---------- 166.20/60.07 TcT has computed the following triangular matrix interpretation. 166.20/60.07 166.20/60.07 [or](x1, x2) = [1] x1 + [1] x2 + [2] 166.20/60.07 166.20/60.07 [and](x1, x2) = [1] x1 + [1] x2 + [2] 166.20/60.07 166.20/60.07 [not](x1) = [1] x1 + [0] 166.20/60.07 166.20/60.07 The order satisfies the following ordering constraints: 166.20/60.07 166.20/60.07 [or(x, x)] = [2] x + [2] 166.20/60.07 > [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [and(x, x)] = [2] x + [2] 166.20/60.07 > [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [not(or(x, y))] = [1] x + [1] y + [2] 166.20/60.07 >= [1] x + [1] y + [2] 166.20/60.07 = [and(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(and(x, y))] = [1] x + [1] y + [2] 166.20/60.07 >= [1] x + [1] y + [2] 166.20/60.07 = [or(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(not(x))] = [1] x + [0] 166.20/60.07 >= [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 166.20/60.07 We return to the main proof. 166.20/60.07 166.20/60.07 We are left with following problem, upon which TcT provides the 166.20/60.07 certificate YES(O(1),O(n^2)). 166.20/60.07 166.20/60.07 Strict Trs: 166.20/60.07 { not(or(x, y)) -> and(not(x), not(y)) 166.20/60.07 , not(and(x, y)) -> or(not(x), not(y)) 166.20/60.07 , not(not(x)) -> x } 166.20/60.07 Weak Trs: 166.20/60.07 { or(x, x) -> x 166.20/60.07 , and(x, x) -> x } 166.20/60.07 Obligation: 166.20/60.07 derivational complexity 166.20/60.07 Answer: 166.20/60.07 YES(O(1),O(n^2)) 166.20/60.07 166.20/60.07 The weightgap principle applies (using the following nonconstant 166.20/60.07 growth matrix-interpretation) 166.20/60.07 166.20/60.07 TcT has computed the following triangular matrix interpretation. 166.20/60.07 Note that the diagonal of the component-wise maxima of 166.20/60.07 interpretation-entries contains no more than 1 non-zero entries. 166.20/60.07 166.20/60.07 [or](x1, x2) = [1] x1 + [1] x2 + [1] 166.20/60.07 166.20/60.07 [and](x1, x2) = [1] x1 + [1] x2 + [0] 166.20/60.07 166.20/60.07 [not](x1) = [1] x1 + [0] 166.20/60.07 166.20/60.07 The order satisfies the following ordering constraints: 166.20/60.07 166.20/60.07 [or(x, x)] = [2] x + [1] 166.20/60.07 > [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [and(x, x)] = [2] x + [0] 166.20/60.07 >= [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [not(or(x, y))] = [1] x + [1] y + [1] 166.20/60.07 > [1] x + [1] y + [0] 166.20/60.07 = [and(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(and(x, y))] = [1] x + [1] y + [0] 166.20/60.07 ? [1] x + [1] y + [1] 166.20/60.07 = [or(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(not(x))] = [1] x + [0] 166.20/60.07 >= [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 166.20/60.07 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 166.20/60.07 166.20/60.07 We are left with following problem, upon which TcT provides the 166.20/60.07 certificate YES(O(1),O(n^2)). 166.20/60.07 166.20/60.07 Strict Trs: 166.20/60.07 { not(and(x, y)) -> or(not(x), not(y)) 166.20/60.07 , not(not(x)) -> x } 166.20/60.07 Weak Trs: 166.20/60.07 { or(x, x) -> x 166.20/60.07 , and(x, x) -> x 166.20/60.07 , not(or(x, y)) -> and(not(x), not(y)) } 166.20/60.07 Obligation: 166.20/60.07 derivational complexity 166.20/60.07 Answer: 166.20/60.07 YES(O(1),O(n^2)) 166.20/60.07 166.20/60.07 The weightgap principle applies (using the following nonconstant 166.20/60.07 growth matrix-interpretation) 166.20/60.07 166.20/60.07 TcT has computed the following triangular matrix interpretation. 166.20/60.07 Note that the diagonal of the component-wise maxima of 166.20/60.07 interpretation-entries contains no more than 1 non-zero entries. 166.20/60.07 166.20/60.07 [or](x1, x2) = [1] x1 + [1] x2 + [2] 166.20/60.07 166.20/60.07 [and](x1, x2) = [1] x1 + [1] x2 + [0] 166.20/60.07 166.20/60.07 [not](x1) = [1] x1 + [2] 166.20/60.07 166.20/60.07 The order satisfies the following ordering constraints: 166.20/60.07 166.20/60.07 [or(x, x)] = [2] x + [2] 166.20/60.07 > [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [and(x, x)] = [2] x + [0] 166.20/60.07 >= [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [not(or(x, y))] = [1] x + [1] y + [4] 166.20/60.07 >= [1] x + [1] y + [4] 166.20/60.07 = [and(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(and(x, y))] = [1] x + [1] y + [2] 166.20/60.07 ? [1] x + [1] y + [6] 166.20/60.07 = [or(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(not(x))] = [1] x + [4] 166.20/60.07 > [1] x + [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 166.20/60.07 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 166.20/60.07 166.20/60.07 We are left with following problem, upon which TcT provides the 166.20/60.07 certificate YES(O(1),O(n^2)). 166.20/60.07 166.20/60.07 Strict Trs: { not(and(x, y)) -> or(not(x), not(y)) } 166.20/60.07 Weak Trs: 166.20/60.07 { or(x, x) -> x 166.20/60.07 , and(x, x) -> x 166.20/60.07 , not(or(x, y)) -> and(not(x), not(y)) 166.20/60.07 , not(not(x)) -> x } 166.20/60.07 Obligation: 166.20/60.07 derivational complexity 166.20/60.07 Answer: 166.20/60.07 YES(O(1),O(n^2)) 166.20/60.07 166.20/60.07 We use the processor 'matrix interpretation of dimension 2' to 166.20/60.07 orient following rules strictly. 166.20/60.07 166.20/60.07 Trs: { not(and(x, y)) -> or(not(x), not(y)) } 166.20/60.07 166.20/60.07 The induced complexity on above rules (modulo remaining rules) is 166.20/60.07 YES(?,O(n^2)) . These rules are moved into the corresponding weak 166.20/60.07 component(s). 166.20/60.07 166.20/60.07 Sub-proof: 166.20/60.07 ---------- 166.20/60.07 TcT has computed the following triangular matrix interpretation. 166.20/60.07 166.20/60.07 [or](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 166.20/60.07 [0 1] [0 1] [1] 166.20/60.07 166.20/60.07 [and](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 166.20/60.07 [0 1] [0 1] [1] 166.20/60.07 166.20/60.07 [not](x1) = [1 1] x1 + [0] 166.20/60.07 [0 1] [0] 166.20/60.07 166.20/60.07 The order satisfies the following ordering constraints: 166.20/60.07 166.20/60.07 [or(x, x)] = [2 0] x + [0] 166.20/60.07 [0 2] [1] 166.20/60.07 >= [1 0] x + [0] 166.20/60.07 [0 1] [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [and(x, x)] = [2 0] x + [0] 166.20/60.07 [0 2] [1] 166.20/60.07 >= [1 0] x + [0] 166.20/60.07 [0 1] [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 [not(or(x, y))] = [1 1] x + [1 1] y + [1] 166.20/60.07 [0 1] [0 1] [1] 166.20/60.07 > [1 1] x + [1 1] y + [0] 166.20/60.07 [0 1] [0 1] [1] 166.20/60.07 = [and(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(and(x, y))] = [1 1] x + [1 1] y + [1] 166.20/60.07 [0 1] [0 1] [1] 166.20/60.07 > [1 1] x + [1 1] y + [0] 166.20/60.07 [0 1] [0 1] [1] 166.20/60.07 = [or(not(x), not(y))] 166.20/60.07 166.20/60.07 [not(not(x))] = [1 2] x + [0] 166.20/60.07 [0 1] [0] 166.20/60.07 >= [1 0] x + [0] 166.20/60.07 [0 1] [0] 166.20/60.07 = [x] 166.20/60.07 166.20/60.07 166.20/60.07 We return to the main proof. 166.20/60.07 166.20/60.07 We are left with following problem, upon which TcT provides the 166.20/60.07 certificate YES(O(1),O(1)). 166.20/60.07 166.20/60.07 Weak Trs: 166.20/60.07 { or(x, x) -> x 166.20/60.07 , and(x, x) -> x 166.20/60.07 , not(or(x, y)) -> and(not(x), not(y)) 166.20/60.07 , not(and(x, y)) -> or(not(x), not(y)) 166.20/60.07 , not(not(x)) -> x } 166.20/60.07 Obligation: 166.20/60.07 derivational complexity 166.20/60.07 Answer: 166.20/60.07 YES(O(1),O(1)) 166.20/60.07 166.20/60.07 Empty rules are trivially bounded 166.20/60.07 166.20/60.07 Hurray, we answered YES(O(1),O(n^2)) 166.20/60.08 EOF