/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). (0) CpxTRS (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (2) TRS for Loop Detection (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] (4) BEST (5) proven lower bound (6) LowerBoundPropagationProof [FINISHED, 0 ms] (7) BOUNDS(n^1, INF) (8) TRS for Loop Detection (9) DecreasingLoopProof [FINISHED, 118 ms] (10) BOUNDS(EXP, INF) ---------------------------------------- (0) Obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: 0(#) -> # +(x, #) -> x +(#, x) -> x +(0(x), 0(y)) -> 0(+(x, y)) +(0(x), 1(y)) -> 1(+(x, y)) +(1(x), 0(y)) -> 1(+(x, y)) +(1(x), 1(y)) -> 0(+(+(x, y), 1(#))) +(+(x, y), z) -> +(x, +(y, z)) -(#, x) -> # -(x, #) -> x -(0(x), 0(y)) -> 0(-(x, y)) -(0(x), 1(y)) -> 1(-(-(x, y), 1(#))) -(1(x), 0(y)) -> 1(-(x, y)) -(1(x), 1(y)) -> 0(-(x, y)) not(true) -> false not(false) -> true if(true, x, y) -> x if(false, x, y) -> y eq(#, #) -> true eq(#, 1(y)) -> false eq(1(x), #) -> false eq(#, 0(y)) -> eq(#, y) eq(0(x), #) -> eq(x, #) eq(1(x), 1(y)) -> eq(x, y) eq(0(x), 1(y)) -> false eq(1(x), 0(y)) -> false eq(0(x), 0(y)) -> eq(x, y) ge(0(x), 0(y)) -> ge(x, y) ge(0(x), 1(y)) -> not(ge(y, x)) ge(1(x), 0(y)) -> ge(x, y) ge(1(x), 1(y)) -> ge(x, y) ge(x, #) -> true ge(#, 0(x)) -> ge(#, x) ge(#, 1(x)) -> false log(x) -> -(log'(x), 1(#)) log'(#) -> # log'(1(x)) -> +(log'(x), 1(#)) log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #) *(#, x) -> # *(0(x), y) -> 0(*(x, y)) *(1(x), y) -> +(0(*(x, y)), y) *(*(x, y), z) -> *(x, *(y, z)) *(x, +(y, z)) -> +(*(x, y), *(x, z)) app(nil, l) -> l app(cons(x, l1), l2) -> cons(x, app(l1, l2)) sum(nil) -> 0(#) sum(cons(x, l)) -> +(x, sum(l)) sum(app(l1, l2)) -> +(sum(l1), sum(l2)) prod(nil) -> 1(#) prod(cons(x, l)) -> *(x, prod(l)) prod(app(l1, l2)) -> *(prod(l1), prod(l2)) mem(x, nil) -> false mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l)) inter(x, nil) -> nil inter(nil, x) -> nil inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3)) inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3)) inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2) inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1) ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2)) ifinter(false, x, l1, l2) -> inter(l1, l2) S is empty. Rewrite Strategy: FULL ---------------------------------------- (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) Transformed a relative TRS into a decreasing-loop problem. ---------------------------------------- (2) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: 0(#) -> # +(x, #) -> x +(#, x) -> x +(0(x), 0(y)) -> 0(+(x, y)) +(0(x), 1(y)) -> 1(+(x, y)) +(1(x), 0(y)) -> 1(+(x, y)) +(1(x), 1(y)) -> 0(+(+(x, y), 1(#))) +(+(x, y), z) -> +(x, +(y, z)) -(#, x) -> # -(x, #) -> x -(0(x), 0(y)) -> 0(-(x, y)) -(0(x), 1(y)) -> 1(-(-(x, y), 1(#))) -(1(x), 0(y)) -> 1(-(x, y)) -(1(x), 1(y)) -> 0(-(x, y)) not(true) -> false not(false) -> true if(true, x, y) -> x if(false, x, y) -> y eq(#, #) -> true eq(#, 1(y)) -> false eq(1(x), #) -> false eq(#, 0(y)) -> eq(#, y) eq(0(x), #) -> eq(x, #) eq(1(x), 1(y)) -> eq(x, y) eq(0(x), 1(y)) -> false eq(1(x), 0(y)) -> false eq(0(x), 0(y)) -> eq(x, y) ge(0(x), 0(y)) -> ge(x, y) ge(0(x), 1(y)) -> not(ge(y, x)) ge(1(x), 0(y)) -> ge(x, y) ge(1(x), 1(y)) -> ge(x, y) ge(x, #) -> true ge(#, 0(x)) -> ge(#, x) ge(#, 1(x)) -> false log(x) -> -(log'(x), 1(#)) log'(#) -> # log'(1(x)) -> +(log'(x), 1(#)) log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #) *(#, x) -> # *(0(x), y) -> 0(*(x, y)) *(1(x), y) -> +(0(*(x, y)), y) *(*(x, y), z) -> *(x, *(y, z)) *(x, +(y, z)) -> +(*(x, y), *(x, z)) app(nil, l) -> l app(cons(x, l1), l2) -> cons(x, app(l1, l2)) sum(nil) -> 0(#) sum(cons(x, l)) -> +(x, sum(l)) sum(app(l1, l2)) -> +(sum(l1), sum(l2)) prod(nil) -> 1(#) prod(cons(x, l)) -> *(x, prod(l)) prod(app(l1, l2)) -> *(prod(l1), prod(l2)) mem(x, nil) -> false mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l)) inter(x, nil) -> nil inter(nil, x) -> nil inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3)) inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3)) inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2) inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1) ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2)) ifinter(false, x, l1, l2) -> inter(l1, l2) S is empty. Rewrite Strategy: FULL ---------------------------------------- (3) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence +(1(x), 1(y)) ->^+ 0(+(+(x, y), 1(#))) gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0]. The pumping substitution is [x / 1(x), y / 1(y)]. The result substitution is [ ]. ---------------------------------------- (4) Complex Obligation (BEST) ---------------------------------------- (5) Obligation: Proved the lower bound n^1 for the following obligation: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: 0(#) -> # +(x, #) -> x +(#, x) -> x +(0(x), 0(y)) -> 0(+(x, y)) +(0(x), 1(y)) -> 1(+(x, y)) +(1(x), 0(y)) -> 1(+(x, y)) +(1(x), 1(y)) -> 0(+(+(x, y), 1(#))) +(+(x, y), z) -> +(x, +(y, z)) -(#, x) -> # -(x, #) -> x -(0(x), 0(y)) -> 0(-(x, y)) -(0(x), 1(y)) -> 1(-(-(x, y), 1(#))) -(1(x), 0(y)) -> 1(-(x, y)) -(1(x), 1(y)) -> 0(-(x, y)) not(true) -> false not(false) -> true if(true, x, y) -> x if(false, x, y) -> y eq(#, #) -> true eq(#, 1(y)) -> false eq(1(x), #) -> false eq(#, 0(y)) -> eq(#, y) eq(0(x), #) -> eq(x, #) eq(1(x), 1(y)) -> eq(x, y) eq(0(x), 1(y)) -> false eq(1(x), 0(y)) -> false eq(0(x), 0(y)) -> eq(x, y) ge(0(x), 0(y)) -> ge(x, y) ge(0(x), 1(y)) -> not(ge(y, x)) ge(1(x), 0(y)) -> ge(x, y) ge(1(x), 1(y)) -> ge(x, y) ge(x, #) -> true ge(#, 0(x)) -> ge(#, x) ge(#, 1(x)) -> false log(x) -> -(log'(x), 1(#)) log'(#) -> # log'(1(x)) -> +(log'(x), 1(#)) log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #) *(#, x) -> # *(0(x), y) -> 0(*(x, y)) *(1(x), y) -> +(0(*(x, y)), y) *(*(x, y), z) -> *(x, *(y, z)) *(x, +(y, z)) -> +(*(x, y), *(x, z)) app(nil, l) -> l app(cons(x, l1), l2) -> cons(x, app(l1, l2)) sum(nil) -> 0(#) sum(cons(x, l)) -> +(x, sum(l)) sum(app(l1, l2)) -> +(sum(l1), sum(l2)) prod(nil) -> 1(#) prod(cons(x, l)) -> *(x, prod(l)) prod(app(l1, l2)) -> *(prod(l1), prod(l2)) mem(x, nil) -> false mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l)) inter(x, nil) -> nil inter(nil, x) -> nil inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3)) inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3)) inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2) inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1) ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2)) ifinter(false, x, l1, l2) -> inter(l1, l2) S is empty. Rewrite Strategy: FULL ---------------------------------------- (6) LowerBoundPropagationProof (FINISHED) Propagated lower bound. ---------------------------------------- (7) BOUNDS(n^1, INF) ---------------------------------------- (8) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(EXP, INF). The TRS R consists of the following rules: 0(#) -> # +(x, #) -> x +(#, x) -> x +(0(x), 0(y)) -> 0(+(x, y)) +(0(x), 1(y)) -> 1(+(x, y)) +(1(x), 0(y)) -> 1(+(x, y)) +(1(x), 1(y)) -> 0(+(+(x, y), 1(#))) +(+(x, y), z) -> +(x, +(y, z)) -(#, x) -> # -(x, #) -> x -(0(x), 0(y)) -> 0(-(x, y)) -(0(x), 1(y)) -> 1(-(-(x, y), 1(#))) -(1(x), 0(y)) -> 1(-(x, y)) -(1(x), 1(y)) -> 0(-(x, y)) not(true) -> false not(false) -> true if(true, x, y) -> x if(false, x, y) -> y eq(#, #) -> true eq(#, 1(y)) -> false eq(1(x), #) -> false eq(#, 0(y)) -> eq(#, y) eq(0(x), #) -> eq(x, #) eq(1(x), 1(y)) -> eq(x, y) eq(0(x), 1(y)) -> false eq(1(x), 0(y)) -> false eq(0(x), 0(y)) -> eq(x, y) ge(0(x), 0(y)) -> ge(x, y) ge(0(x), 1(y)) -> not(ge(y, x)) ge(1(x), 0(y)) -> ge(x, y) ge(1(x), 1(y)) -> ge(x, y) ge(x, #) -> true ge(#, 0(x)) -> ge(#, x) ge(#, 1(x)) -> false log(x) -> -(log'(x), 1(#)) log'(#) -> # log'(1(x)) -> +(log'(x), 1(#)) log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #) *(#, x) -> # *(0(x), y) -> 0(*(x, y)) *(1(x), y) -> +(0(*(x, y)), y) *(*(x, y), z) -> *(x, *(y, z)) *(x, +(y, z)) -> +(*(x, y), *(x, z)) app(nil, l) -> l app(cons(x, l1), l2) -> cons(x, app(l1, l2)) sum(nil) -> 0(#) sum(cons(x, l)) -> +(x, sum(l)) sum(app(l1, l2)) -> +(sum(l1), sum(l2)) prod(nil) -> 1(#) prod(cons(x, l)) -> *(x, prod(l)) prod(app(l1, l2)) -> *(prod(l1), prod(l2)) mem(x, nil) -> false mem(x, cons(y, l)) -> if(eq(x, y), true, mem(x, l)) inter(x, nil) -> nil inter(nil, x) -> nil inter(app(l1, l2), l3) -> app(inter(l1, l3), inter(l2, l3)) inter(l1, app(l2, l3)) -> app(inter(l1, l2), inter(l1, l3)) inter(cons(x, l1), l2) -> ifinter(mem(x, l2), x, l1, l2) inter(l1, cons(x, l2)) -> ifinter(mem(x, l1), x, l2, l1) ifinter(true, x, l1, l2) -> cons(x, inter(l1, l2)) ifinter(false, x, l1, l2) -> inter(l1, l2) S is empty. Rewrite Strategy: FULL ---------------------------------------- (9) DecreasingLoopProof (FINISHED) The following loop(s) give(s) rise to the lower bound EXP: The rewrite sequence prod(cons(1(x1_0), l)) ->^+ +(0(*(x1_0, prod(l))), prod(l)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,1]. The pumping substitution is [l / cons(1(x1_0), l)]. The result substitution is [ ]. The rewrite sequence prod(cons(1(x1_0), l)) ->^+ +(0(*(x1_0, prod(l))), prod(l)) gives rise to a decreasing loop by considering the right hand sides subterm at position [1]. The pumping substitution is [l / cons(1(x1_0), l)]. The result substitution is [ ]. ---------------------------------------- (10) BOUNDS(EXP, INF)