YES proof of SK90_2.39.trs # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 70 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 0 ms] (8) QTRS (9) RisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: rev(nil) -> nil rev(.(x, y)) -> ++(rev(y), .(x, nil)) car(.(x, y)) -> x cdr(.(x, y)) -> y null(nil) -> true null(.(x, y)) -> false ++(nil, y) -> y ++(.(x, y), z) -> .(x, ++(y, z)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(++(x_1, x_2)) = x_1 + x_2 POL(.(x_1, x_2)) = x_1 + x_2 POL(car(x_1)) = 1 + x_1 POL(cdr(x_1)) = 2 + x_1 POL(false) = 0 POL(nil) = 0 POL(null(x_1)) = 1 + x_1 POL(rev(x_1)) = 2*x_1 POL(true) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: car(.(x, y)) -> x cdr(.(x, y)) -> y null(nil) -> true null(.(x, y)) -> false ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: rev(nil) -> nil rev(.(x, y)) -> ++(rev(y), .(x, nil)) ++(nil, y) -> y ++(.(x, y), z) -> .(x, ++(y, z)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(++(x_1, x_2)) = 1 + x_1 + x_2 POL(.(x_1, x_2)) = 1 + x_1 + x_2 POL(nil) = 0 POL(rev(x_1)) = 2 + 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: rev(nil) -> nil ++(nil, y) -> y ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: rev(.(x, y)) -> ++(rev(y), .(x, nil)) ++(.(x, y), z) -> .(x, ++(y, z)) Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(++(x_1, x_2)) = x_1 + x_2 POL(.(x_1, x_2)) = 1 + x_1 + x_2 POL(nil) = 0 POL(rev(x_1)) = 2 + 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: rev(.(x, y)) -> ++(rev(y), .(x, nil)) ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: ++(.(x, y), z) -> .(x, ++(y, z)) Q is empty. ---------------------------------------- (7) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:++_2 > ._2 and weight map: ._2=0 ++_2=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: ++(.(x, y), z) -> .(x, ++(y, z)) ---------------------------------------- (8) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (9) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (10) YES