YES(O(1),O(n^2)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N) , plus(N, 0()) -> U11(isNat(N), N) , and(tt(), X) -> activate(X) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) Arguments of following rules are not normal-forms: { plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N) , plus(N, 0()) -> U11(isNat(N), N) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , and(tt(), X) -> activate(X) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [1] x1 + [1] x2 + [7] [tt] = [7] [activate](x1) = [1] x1 + [3] [U21](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [7] [s](x1) = [1] x1 + [1] [plus](x1, x2) = [1] x1 + [1] x2 + [7] [and](x1, x2) = [1] x1 + [1] x2 + [3] [isNat](x1) = [1] x1 + [3] [n__0] = [3] [n__plus](x1, x2) = [1] x1 + [1] x2 + [7] [n__isNat](x1) = [1] x1 + [3] [n__s](x1) = [1] x1 + [7] [0] = [5] The following symbols are considered usable {U11, activate, U21, s, plus, and, isNat, 0} The order satisfies the following ordering constraints: [U11(tt(), N)] = [1] N + [14] > [1] N + [3] = [activate(N)] [activate(X)] = [1] X + [3] > [1] X + [0] = [X] [activate(n__0())] = [6] > [5] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [10] ? [1] X1 + [1] X2 + [13] = [plus(activate(X1), activate(X2))] [activate(n__isNat(X))] = [1] X + [6] > [1] X + [3] = [isNat(X)] [activate(n__s(X))] = [1] X + [10] > [1] X + [4] = [s(activate(X))] [U21(tt(), M, N)] = [1] N + [1] M + [14] >= [1] N + [1] M + [14] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [1] ? [1] X + [7] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [7] >= [1] X1 + [1] X2 + [7] = [n__plus(X1, X2)] [and(tt(), X)] = [1] X + [10] > [1] X + [3] = [activate(X)] [isNat(X)] = [1] X + [3] >= [1] X + [3] = [n__isNat(X)] [isNat(n__0())] = [6] ? [7] = [tt()] [isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [10] ? [1] V1 + [1] V2 + [15] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [1] V1 + [10] > [1] V1 + [6] = [isNat(activate(V1))] [0()] = [5] > [3] = [n__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 YES(O(1),O(n^2)). Strict Trs: { activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , and(tt(), X) -> activate(X) , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [1] x1 + [1] x2 + [7] [tt] = [0] [activate](x1) = [1] x1 + [0] [U21](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [7] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [and](x1, x2) = [1] x1 + [1] x2 + [0] [isNat](x1) = [1] x1 + [1] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [0] [n__isNat](x1) = [1] x1 + [7] [n__s](x1) = [1] x1 + [3] [0] = [0] The following symbols are considered usable {U11, activate, U21, s, plus, and, isNat, 0} The order satisfies the following ordering constraints: [U11(tt(), N)] = [1] N + [7] > [1] N + [0] = [activate(N)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [0] >= [0] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [plus(activate(X1), activate(X2))] [activate(n__isNat(X))] = [1] X + [7] > [1] X + [1] = [isNat(X)] [activate(n__s(X))] = [1] X + [3] > [1] X + [0] = [s(activate(X))] [U21(tt(), M, N)] = [1] N + [1] M + [7] > [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] ? [1] X + [3] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__plus(X1, X2)] [and(tt(), X)] = [1] X + [0] >= [1] X + [0] = [activate(X)] [isNat(X)] = [1] X + [1] ? [1] X + [7] = [n__isNat(X)] [isNat(n__0())] = [1] > [0] = [tt()] [isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [1] ? [1] V1 + [1] V2 + [8] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [1] V1 + [4] > [1] V1 + [1] = [isNat(activate(V1))] [0()] = [0] >= [0] = [n__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 YES(O(1),O(n^2)). Strict Trs: { activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , isNat(X) -> n__isNat(X) , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , and(tt(), X) -> activate(X) , isNat(n__0()) -> tt() , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [1] x1 + [1] x2 + [7] [tt] = [4] [activate](x1) = [1] x1 + [0] [U21](x1, x2, x3) = [1] x2 + [1] x3 + [7] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [and](x1, x2) = [1] x1 + [1] x2 + [0] [isNat](x1) = [1] x1 + [4] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [4] [n__isNat](x1) = [1] x1 + [4] [n__s](x1) = [1] x1 + [0] [0] = [0] The following symbols are considered usable {U11, activate, U21, s, plus, and, isNat, 0} The order satisfies the following ordering constraints: [U11(tt(), N)] = [1] N + [11] > [1] N + [0] = [activate(N)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [0] >= [0] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [4] > [1] X1 + [1] X2 + [0] = [plus(activate(X1), activate(X2))] [activate(n__isNat(X))] = [1] X + [4] >= [1] X + [4] = [isNat(X)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(activate(X))] [U21(tt(), M, N)] = [1] N + [1] M + [7] > [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] ? [1] X1 + [1] X2 + [4] = [n__plus(X1, X2)] [and(tt(), X)] = [1] X + [4] > [1] X + [0] = [activate(X)] [isNat(X)] = [1] X + [4] >= [1] X + [4] = [n__isNat(X)] [isNat(n__0())] = [4] >= [4] = [tt()] [isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [8] >= [1] V1 + [1] V2 + [8] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [1] V1 + [4] >= [1] V1 + [4] = [isNat(activate(V1))] [0()] = [0] >= [0] = [n__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 YES(O(1),O(n^2)). Strict Trs: { s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , isNat(X) -> n__isNat(X) , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , and(tt(), X) -> activate(X) , isNat(n__0()) -> tt() , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [1] x1 + [1] x2 + [7] [tt] = [1] [activate](x1) = [1] x1 + [1] [U21](x1, x2, x3) = [1] x2 + [1] x3 + [7] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [2] [and](x1, x2) = [1] x1 + [1] x2 + [0] [isNat](x1) = [1] x1 + [4] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [4] [n__isNat](x1) = [1] x1 + [3] [n__s](x1) = [1] x1 + [4] [0] = [1] The following symbols are considered usable {U11, activate, U21, s, plus, and, isNat, 0} The order satisfies the following ordering constraints: [U11(tt(), N)] = [1] N + [8] > [1] N + [1] = [activate(N)] [activate(X)] = [1] X + [1] > [1] X + [0] = [X] [activate(n__0())] = [1] >= [1] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [5] > [1] X1 + [1] X2 + [4] = [plus(activate(X1), activate(X2))] [activate(n__isNat(X))] = [1] X + [4] >= [1] X + [4] = [isNat(X)] [activate(n__s(X))] = [1] X + [5] > [1] X + [1] = [s(activate(X))] [U21(tt(), M, N)] = [1] N + [1] M + [7] > [1] N + [1] M + [4] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] ? [1] X + [4] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [2] ? [1] X1 + [1] X2 + [4] = [n__plus(X1, X2)] [and(tt(), X)] = [1] X + [1] >= [1] X + [1] = [activate(X)] [isNat(X)] = [1] X + [4] > [1] X + [3] = [n__isNat(X)] [isNat(n__0())] = [4] > [1] = [tt()] [isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [8] ? [1] V1 + [1] V2 + [9] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [1] V1 + [8] > [1] V1 + [5] = [isNat(activate(V1))] [0()] = [1] > [0] = [n__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 YES(O(1),O(n^2)). Strict Trs: { s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , and(tt(), X) -> activate(X) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [1] x1 + [1] x2 + [7] [tt] = [0] [activate](x1) = [1] x1 + [0] [U21](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [3] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [and](x1, x2) = [1] x1 + [1] x2 + [0] [isNat](x1) = [1] x1 + [0] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [4] [n__isNat](x1) = [1] x1 + [0] [n__s](x1) = [1] x1 + [0] [0] = [0] The following symbols are considered usable {U11, activate, U21, s, plus, and, isNat, 0} The order satisfies the following ordering constraints: [U11(tt(), N)] = [1] N + [7] > [1] N + [0] = [activate(N)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [0] >= [0] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [4] > [1] X1 + [1] X2 + [0] = [plus(activate(X1), activate(X2))] [activate(n__isNat(X))] = [1] X + [0] >= [1] X + [0] = [isNat(X)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(activate(X))] [U21(tt(), M, N)] = [1] N + [1] M + [3] > [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] ? [1] X1 + [1] X2 + [4] = [n__plus(X1, X2)] [and(tt(), X)] = [1] X + [0] >= [1] X + [0] = [activate(X)] [isNat(X)] = [1] X + [0] >= [1] X + [0] = [n__isNat(X)] [isNat(n__0())] = [0] >= [0] = [tt()] [isNat(n__plus(V1, V2))] = [1] V1 + [1] V2 + [4] > [1] V1 + [1] V2 + [0] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [1] V1 + [0] >= [1] V1 + [0] = [isNat(activate(V1))] [0()] = [0] >= [0] = [n__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 YES(O(1),O(n^2)). Strict Trs: { s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , and(tt(), X) -> activate(X) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { s(X) -> n__s(X) } 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: Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [7 4] x1 + [7 7] x2 + [7] [7 7] [7 7] [7] [tt] = [0] [0] [activate](x1) = [1 1] x1 + [0] [0 1] [0] [U21](x1, x2, x3) = [7 4] x1 + [7 7] x2 + [7 7] x3 + [7] [7 7] [7 7] [7 7] [7] [s](x1) = [1 1] x1 + [4] [0 1] [4] [plus](x1, x2) = [1 1] x1 + [1 1] x2 + [0] [0 1] [0 1] [0] [and](x1, x2) = [1 0] x1 + [1 1] x2 + [0] [0 0] [0 2] [0] [isNat](x1) = [1 0] x1 + [0] [0 0] [0] [n__0] = [0] [0] [n__plus](x1, x2) = [1 1] x1 + [1 1] x2 + [0] [0 1] [0 1] [0] [n__isNat](x1) = [1 0] x1 + [0] [0 0] [0] [n__s](x1) = [1 1] x1 + [0] [0 1] [4] [0] = [0] [0] The following symbols are considered usable {U11, activate, U21, s, plus, and, isNat, 0} The order satisfies the following ordering constraints: [U11(tt(), N)] = [7 7] N + [7] [7 7] [7] > [1 1] N + [0] [0 1] [0] = [activate(N)] [activate(X)] = [1 1] X + [0] [0 1] [0] >= [1 0] X + [0] [0 1] [0] = [X] [activate(n__0())] = [0] [0] >= [0] [0] = [0()] [activate(n__plus(X1, X2))] = [1 2] X1 + [1 2] X2 + [0] [0 1] [0 1] [0] >= [1 2] X1 + [1 2] X2 + [0] [0 1] [0 1] [0] = [plus(activate(X1), activate(X2))] [activate(n__isNat(X))] = [1 0] X + [0] [0 0] [0] >= [1 0] X + [0] [0 0] [0] = [isNat(X)] [activate(n__s(X))] = [1 2] X + [4] [0 1] [4] >= [1 2] X + [4] [0 1] [4] = [s(activate(X))] [U21(tt(), M, N)] = [7 7] N + [7 7] M + [7] [7 7] [7 7] [7] > [1 3] N + [1 3] M + [4] [0 1] [0 1] [4] = [s(plus(activate(N), activate(M)))] [s(X)] = [1 1] X + [4] [0 1] [4] > [1 1] X + [0] [0 1] [4] = [n__s(X)] [plus(X1, X2)] = [1 1] X1 + [1 1] X2 + [0] [0 1] [0 1] [0] >= [1 1] X1 + [1 1] X2 + [0] [0 1] [0 1] [0] = [n__plus(X1, X2)] [and(tt(), X)] = [1 1] X + [0] [0 2] [0] >= [1 1] X + [0] [0 1] [0] = [activate(X)] [isNat(X)] = [1 0] X + [0] [0 0] [0] >= [1 0] X + [0] [0 0] [0] = [n__isNat(X)] [isNat(n__0())] = [0] [0] >= [0] [0] = [tt()] [isNat(n__plus(V1, V2))] = [1 1] V1 + [1 1] V2 + [0] [0 0] [0 0] [0] >= [1 1] V1 + [1 1] V2 + [0] [0 0] [0 0] [0] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [1 1] V1 + [0] [0 0] [0] >= [1 1] V1 + [0] [0 0] [0] = [isNat(activate(V1))] [0()] = [0] [0] >= [0] [0] = [n__0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { plus(X1, X2) -> n__plus(X1, X2) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , and(tt(), X) -> activate(X) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { plus(X1, X2) -> n__plus(X1, X2) } 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: Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(and) = {1, 2}, Uargs(isNat) = {1}, Uargs(n__isNat) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [0 4] x1 + [7 7] x2 + [7] [0 7] [7 7] [7] [tt] = [4] [0] [activate](x1) = [1 1] x1 + [0] [0 1] [0] [U21](x1, x2, x3) = [2 4] x1 + [7 7] x2 + [7 7] x3 + [7] [0 7] [7 7] [7 7] [7] [s](x1) = [1 4] x1 + [0] [0 1] [0] [plus](x1, x2) = [1 1] x1 + [1 1] x2 + [1] [0 1] [0 1] [2] [and](x1, x2) = [1 0] x1 + [1 2] x2 + [0] [0 0] [0 4] [0] [isNat](x1) = [1 0] x1 + [0] [0 0] [0] [n__0] = [7] [1] [n__plus](x1, x2) = [1 1] x1 + [1 1] x2 + [0] [0 1] [0 1] [2] [n__isNat](x1) = [1 0] x1 + [0] [0 0] [0] [n__s](x1) = [1 4] x1 + [0] [0 1] [0] [0] = [7] [1] The following symbols are considered usable {U11, activate, U21, s, plus, and, isNat, 0} The order satisfies the following ordering constraints: [U11(tt(), N)] = [7 7] N + [7] [7 7] [7] > [1 1] N + [0] [0 1] [0] = [activate(N)] [activate(X)] = [1 1] X + [0] [0 1] [0] >= [1 0] X + [0] [0 1] [0] = [X] [activate(n__0())] = [8] [1] > [7] [1] = [0()] [activate(n__plus(X1, X2))] = [1 2] X1 + [1 2] X2 + [2] [0 1] [0 1] [2] > [1 2] X1 + [1 2] X2 + [1] [0 1] [0 1] [2] = [plus(activate(X1), activate(X2))] [activate(n__isNat(X))] = [1 0] X + [0] [0 0] [0] >= [1 0] X + [0] [0 0] [0] = [isNat(X)] [activate(n__s(X))] = [1 5] X + [0] [0 1] [0] >= [1 5] X + [0] [0 1] [0] = [s(activate(X))] [U21(tt(), M, N)] = [7 7] N + [7 7] M + [15] [7 7] [7 7] [7] > [1 6] N + [1 6] M + [9] [0 1] [0 1] [2] = [s(plus(activate(N), activate(M)))] [s(X)] = [1 4] X + [0] [0 1] [0] >= [1 4] X + [0] [0 1] [0] = [n__s(X)] [plus(X1, X2)] = [1 1] X1 + [1 1] X2 + [1] [0 1] [0 1] [2] > [1 1] X1 + [1 1] X2 + [0] [0 1] [0 1] [2] = [n__plus(X1, X2)] [and(tt(), X)] = [1 2] X + [4] [0 4] [0] > [1 1] X + [0] [0 1] [0] = [activate(X)] [isNat(X)] = [1 0] X + [0] [0 0] [0] >= [1 0] X + [0] [0 0] [0] = [n__isNat(X)] [isNat(n__0())] = [7] [0] > [4] [0] = [tt()] [isNat(n__plus(V1, V2))] = [1 1] V1 + [1 1] V2 + [0] [0 0] [0 0] [0] >= [1 1] V1 + [1 1] V2 + [0] [0 0] [0 0] [0] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [1 4] V1 + [0] [0 0] [0] >= [1 1] V1 + [0] [0 0] [0] = [isNat(activate(V1))] [0()] = [7] [1] >= [7] [1] = [n__0()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(activate(X)) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , and(tt(), X) -> activate(X) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> and(isNat(activate(V1)), n__isNat(activate(V2))) , isNat(n__s(V1)) -> isNat(activate(V1)) , 0() -> n__0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^2))