3.99/1.79 WORST_CASE(NON_POLY, ?) 4.23/1.81 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.23/1.81 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.23/1.81 4.23/1.81 4.23/1.81 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 4.23/1.81 4.23/1.81 (0) CpxTRS 4.23/1.81 (1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] 4.23/1.81 (2) TRS for Loop Detection 4.23/1.81 (3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] 4.23/1.81 (4) BEST 4.23/1.81 (5) proven lower bound 4.23/1.81 (6) LowerBoundPropagationProof [FINISHED, 0 ms] 4.23/1.81 (7) BOUNDS(n^1, INF) 4.23/1.81 (8) TRS for Loop Detection 4.23/1.81 (9) InfiniteLowerBoundProof [FINISHED, 117 ms] 4.23/1.81 (10) BOUNDS(INF, INF) 4.23/1.81 4.23/1.81 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (0) 4.23/1.81 Obligation: 4.23/1.81 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 4.23/1.81 4.23/1.81 4.23/1.81 The TRS R consists of the following rules: 4.23/1.81 4.23/1.81 incr(nil) -> nil 4.23/1.81 incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) 4.23/1.81 adx(nil) -> nil 4.23/1.81 adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) 4.23/1.81 nats -> adx(zeros) 4.23/1.81 zeros -> cons(0, n__zeros) 4.23/1.81 head(cons(X, L)) -> X 4.23/1.81 tail(cons(X, L)) -> activate(L) 4.23/1.81 incr(X) -> n__incr(X) 4.23/1.81 adx(X) -> n__adx(X) 4.23/1.81 zeros -> n__zeros 4.23/1.81 activate(n__incr(X)) -> incr(X) 4.23/1.81 activate(n__adx(X)) -> adx(X) 4.23/1.81 activate(n__zeros) -> zeros 4.23/1.81 activate(X) -> X 4.23/1.81 4.23/1.81 S is empty. 4.23/1.81 Rewrite Strategy: INNERMOST 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) 4.23/1.81 Transformed a relative TRS into a decreasing-loop problem. 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (2) 4.23/1.81 Obligation: 4.23/1.81 Analyzing the following TRS for decreasing loops: 4.23/1.81 4.23/1.81 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 4.23/1.81 4.23/1.81 4.23/1.81 The TRS R consists of the following rules: 4.23/1.81 4.23/1.81 incr(nil) -> nil 4.23/1.81 incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) 4.23/1.81 adx(nil) -> nil 4.23/1.81 adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) 4.23/1.81 nats -> adx(zeros) 4.23/1.81 zeros -> cons(0, n__zeros) 4.23/1.81 head(cons(X, L)) -> X 4.23/1.81 tail(cons(X, L)) -> activate(L) 4.23/1.81 incr(X) -> n__incr(X) 4.23/1.81 adx(X) -> n__adx(X) 4.23/1.81 zeros -> n__zeros 4.23/1.81 activate(n__incr(X)) -> incr(X) 4.23/1.81 activate(n__adx(X)) -> adx(X) 4.23/1.81 activate(n__zeros) -> zeros 4.23/1.81 activate(X) -> X 4.23/1.81 4.23/1.81 S is empty. 4.23/1.81 Rewrite Strategy: INNERMOST 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (3) DecreasingLoopProof (LOWER BOUND(ID)) 4.23/1.81 The following loop(s) give(s) rise to the lower bound Omega(n^1): 4.23/1.81 4.23/1.81 The rewrite sequence 4.23/1.81 4.23/1.81 activate(n__incr(cons(X1_0, L2_0))) ->^+ cons(s(X1_0), n__incr(activate(L2_0))) 4.23/1.81 4.23/1.81 gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0]. 4.23/1.81 4.23/1.81 The pumping substitution is [L2_0 / n__incr(cons(X1_0, L2_0))]. 4.23/1.81 4.23/1.81 The result substitution is [ ]. 4.23/1.81 4.23/1.81 4.23/1.81 4.23/1.81 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (4) 4.23/1.81 Complex Obligation (BEST) 4.23/1.81 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (5) 4.23/1.81 Obligation: 4.23/1.81 Proved the lower bound n^1 for the following obligation: 4.23/1.81 4.23/1.81 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 4.23/1.81 4.23/1.81 4.23/1.81 The TRS R consists of the following rules: 4.23/1.81 4.23/1.81 incr(nil) -> nil 4.23/1.81 incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) 4.23/1.81 adx(nil) -> nil 4.23/1.81 adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) 4.23/1.81 nats -> adx(zeros) 4.23/1.81 zeros -> cons(0, n__zeros) 4.23/1.81 head(cons(X, L)) -> X 4.23/1.81 tail(cons(X, L)) -> activate(L) 4.23/1.81 incr(X) -> n__incr(X) 4.23/1.81 adx(X) -> n__adx(X) 4.23/1.81 zeros -> n__zeros 4.23/1.81 activate(n__incr(X)) -> incr(X) 4.23/1.81 activate(n__adx(X)) -> adx(X) 4.23/1.81 activate(n__zeros) -> zeros 4.23/1.81 activate(X) -> X 4.23/1.81 4.23/1.81 S is empty. 4.23/1.81 Rewrite Strategy: INNERMOST 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (6) LowerBoundPropagationProof (FINISHED) 4.23/1.81 Propagated lower bound. 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (7) 4.23/1.81 BOUNDS(n^1, INF) 4.23/1.81 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (8) 4.23/1.81 Obligation: 4.23/1.81 Analyzing the following TRS for decreasing loops: 4.23/1.81 4.23/1.81 The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(INF, INF). 4.23/1.81 4.23/1.81 4.23/1.81 The TRS R consists of the following rules: 4.23/1.81 4.23/1.81 incr(nil) -> nil 4.23/1.81 incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) 4.23/1.81 adx(nil) -> nil 4.23/1.81 adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) 4.23/1.81 nats -> adx(zeros) 4.23/1.81 zeros -> cons(0, n__zeros) 4.23/1.81 head(cons(X, L)) -> X 4.23/1.81 tail(cons(X, L)) -> activate(L) 4.23/1.81 incr(X) -> n__incr(X) 4.23/1.81 adx(X) -> n__adx(X) 4.23/1.81 zeros -> n__zeros 4.23/1.81 activate(n__incr(X)) -> incr(X) 4.23/1.81 activate(n__adx(X)) -> adx(X) 4.23/1.81 activate(n__zeros) -> zeros 4.23/1.81 activate(X) -> X 4.23/1.81 4.23/1.81 S is empty. 4.23/1.81 Rewrite Strategy: INNERMOST 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (9) InfiniteLowerBoundProof (FINISHED) 4.23/1.81 The following loop proves infinite runtime complexity: 4.23/1.81 4.23/1.81 The rewrite sequence 4.23/1.81 4.23/1.81 adx(cons(X, n__zeros)) ->^+ cons(s(X), n__incr(adx(cons(0, n__zeros)))) 4.23/1.81 4.23/1.81 gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0]. 4.23/1.81 4.23/1.81 The pumping substitution is [ ]. 4.23/1.81 4.23/1.81 The result substitution is [X / 0]. 4.23/1.81 4.23/1.81 4.23/1.81 4.23/1.81 4.23/1.81 ---------------------------------------- 4.23/1.81 4.23/1.81 (10) 4.23/1.81 BOUNDS(INF, INF) 4.28/2.41 EOF