KILLED
Runtime Complexity (innermost) proof of /tmp/tmpAnCDjB/4.05.xml
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).
0 CpxTRS
↳1 DecreasingLoopProof (⇔, 391 ms)
↳2 BOUNDS(n^1, INF)
↳3 RenamingProof (⇔, 0 ms)
↳4 CpxRelTRS
↳5 TypeInferenceProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 typed CpxTrs
↳7 OrderProof (LOWER BOUND(ID), 0 ms)
↳8 typed CpxTrs
(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, 1) → x
*(1, y) → y
Rewrite Strategy: INNERMOST
(1) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(n1):
The rewrite sequence
*(x, +(y, z)) →+ +(*(x, y), *(x, z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [y / +(y, z)].
The result substitution is [ ].
(2) BOUNDS(n^1, INF)
(3) RenamingProof (EQUIVALENT transformation)
Renamed function symbols to avoid clashes with predefined symbol.
(4) Obligation:
Runtime Complexity Relative TRS:
The TRS R consists of the following rules:
*'(x, +'(y, z)) → +'(*'(x, y), *'(x, z))
*'(+'(x, y), z) → +'(*'(x, z), *'(y, z))
*'(x, 1') → x
*'(1', y) → y
S is empty.
Rewrite Strategy: INNERMOST
(5) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)
Infered types.
(6) Obligation:
Innermost TRS:
Rules:
*'(x, +'(y, z)) → +'(*'(x, y), *'(x, z))
*'(+'(x, y), z) → +'(*'(x, z), *'(y, z))
*'(x, 1') → x
*'(1', y) → y
Types:
*' :: +':1' → +':1' → +':1'
+' :: +':1' → +':1' → +':1'
1' :: +':1'
hole_+':1'1_0 :: +':1'
gen_+':1'2_0 :: Nat → +':1'
(7) OrderProof (LOWER BOUND(ID) transformation)
Heuristically decided to analyse the following defined symbols:
*'
(8) Obligation:
Innermost TRS:
Rules:
*'(x, +'(y, z)) → +'(*'(x, y), *'(x, z))
*'(+'(x, y), z) → +'(*'(x, z), *'(y, z))
*'(x, 1') → x
*'(1', y) → y
Types:
*' :: +':1' → +':1' → +':1'
+' :: +':1' → +':1' → +':1'
1' :: +':1'
hole_+':1'1_0 :: +':1'
gen_+':1'2_0 :: Nat → +':1'
Generator Equations:
gen_+':1'2_0(0) ⇔ 1'
gen_+':1'2_0(+(x, 1)) ⇔ +'(1', gen_+':1'2_0(x))
The following defined symbols remain to be analysed:
*'