/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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WORST_CASE(Omega(n^1), O(n^2))
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
    (3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (4) CdtProblem
    (5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 81 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 131 ms]
    (10) CdtProblem
    (11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 76 ms]
    (12) CdtProblem
    (13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 95 ms]
    (14) CdtProblem
    (15) CdtKnowledgeProof [FINISHED, 0 ms]
    (16) BOUNDS(1, 1)
(17) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
(18) CpxTRS
    (19) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
    (20) typed CpxTrs
    (21) OrderProof [LOWER BOUND(ID), 0 ms]
    (22) typed CpxTrs
    (23) RewriteLemmaProof [LOWER BOUND(ID), 315 ms]
    (24) BEST
        (25) proven lower bound
            (26) LowerBoundPropagationProof [FINISHED, 0 ms]
            (27) BOUNDS(n^1, INF)
        (28) typed CpxTrs
            (29) RewriteLemmaProof [LOWER BOUND(ID), 37 ms]
            (30) typed CpxTrs


----------------------------------------

(0)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2).


The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(X)) -> false
   eq(s(X), 0) -> false
   eq(s(X), s(Y)) -> eq(X, Y)
   rm(N, nil) -> nil
   rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
   ifrm(true, N, add(M, X)) -> rm(N, X)
   ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
   purge(nil) -> nil
   purge(add(N, X)) -> add(N, purge(rm(N, X)))

S is empty.
Rewrite Strategy: INNERMOST
----------------------------------------

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   rm(z0, nil) -> nil
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
   purge(nil) -> nil
   purge(add(z0, z1)) -> add(z0, purge(rm(z0, z1)))
Tuples:
   EQ(0, 0) -> c
   EQ(0, s(z0)) -> c1
   EQ(s(z0), 0) -> c2
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, nil) -> c4
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(nil) -> c8
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
S tuples:
   EQ(0, 0) -> c
   EQ(0, s(z0)) -> c1
   EQ(s(z0), 0) -> c2
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, nil) -> c4
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(nil) -> c8
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
K tuples:none
Defined Rule Symbols:   eq_2, rm_2, ifrm_3, purge_1

Defined Pair Symbols:   EQ_2, RM_2, IFRM_3, PURGE_1

Compound Symbols:   c, c1, c2, c3_1, c4, c5_2, c6_1, c7_1, c8, c9_2


----------------------------------------

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 5 trailing nodes:
   EQ(0, 0) -> c
   PURGE(nil) -> c8
   EQ(s(z0), 0) -> c2
   EQ(0, s(z0)) -> c1
   RM(z0, nil) -> c4

----------------------------------------

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   rm(z0, nil) -> nil
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
   purge(nil) -> nil
   purge(add(z0, z1)) -> add(z0, purge(rm(z0, z1)))
Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
S tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
K tuples:none
Defined Rule Symbols:   eq_2, rm_2, ifrm_3, purge_1

Defined Pair Symbols:   EQ_2, RM_2, IFRM_3, PURGE_1

Compound Symbols:   c3_1, c5_2, c6_1, c7_1, c9_2


----------------------------------------

(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   purge(nil) -> nil
   purge(add(z0, z1)) -> add(z0, purge(rm(z0, z1)))

----------------------------------------

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   rm(z0, nil) -> nil
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
S tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
K tuples:none
Defined Rule Symbols:   eq_2, rm_2, ifrm_3

Defined Pair Symbols:   EQ_2, RM_2, IFRM_3, PURGE_1

Compound Symbols:   c3_1, c5_2, c6_1, c7_1, c9_2


----------------------------------------

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
We considered the (Usable) Rules:
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
   rm(z0, nil) -> nil
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
And the Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(EQ(x_1, x_2)) = 0
   POL(IFRM(x_1, x_2, x_3)) = 0
   POL(PURGE(x_1)) = x_1
   POL(RM(x_1, x_2)) = 0
   POL(add(x_1, x_2)) = [1] + x_1 + x_2
   POL(c3(x_1)) = x_1
   POL(c5(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(c9(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = [1] + x_1 + x_2
   POL(false) = [1]
   POL(ifrm(x_1, x_2, x_3)) = x_2 + x_3
   POL(nil) = [1]
   POL(rm(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = [1] + x_1
   POL(true) = [1]

----------------------------------------

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   rm(z0, nil) -> nil
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
S tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
K tuples:
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
Defined Rule Symbols:   eq_2, rm_2, ifrm_3

Defined Pair Symbols:   EQ_2, RM_2, IFRM_3, PURGE_1

Compound Symbols:   c3_1, c5_2, c6_1, c7_1, c9_2


----------------------------------------

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
We considered the (Usable) Rules:
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
   rm(z0, nil) -> nil
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
And the Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(EQ(x_1, x_2)) = x_1
   POL(IFRM(x_1, x_2, x_3)) = [2]x_2*x_3
   POL(PURGE(x_1)) = [2]x_1^2
   POL(RM(x_1, x_2)) = [2]x_1 + [2]x_1*x_2
   POL(add(x_1, x_2)) = [2] + x_1 + x_2
   POL(c3(x_1)) = x_1
   POL(c5(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(c9(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = 0
   POL(false) = 0
   POL(ifrm(x_1, x_2, x_3)) = x_3
   POL(nil) = 0
   POL(rm(x_1, x_2)) = x_2
   POL(s(x_1)) = [2] + x_1
   POL(true) = 0

----------------------------------------

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   rm(z0, nil) -> nil
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
S tuples:
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
K tuples:
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
Defined Rule Symbols:   eq_2, rm_2, ifrm_3

Defined Pair Symbols:   EQ_2, RM_2, IFRM_3, PURGE_1

Compound Symbols:   c3_1, c5_2, c6_1, c7_1, c9_2


----------------------------------------

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
We considered the (Usable) Rules:
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   eq(0, 0) -> true
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
   rm(z0, nil) -> nil
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
And the Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = [1]
   POL(EQ(x_1, x_2)) = 0
   POL(IFRM(x_1, x_2, x_3)) = x_1 + [2]x_2*x_3
   POL(PURGE(x_1)) = x_1^2
   POL(RM(x_1, x_2)) = [2]x_1 + [2]x_1*x_2
   POL(add(x_1, x_2)) = [1] + x_1 + x_2
   POL(c3(x_1)) = x_1
   POL(c5(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(c9(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = x_1
   POL(false) = [1]
   POL(ifrm(x_1, x_2, x_3)) = x_3
   POL(nil) = 0
   POL(rm(x_1, x_2)) = x_2
   POL(s(x_1)) = [1] + x_1
   POL(true) = 0

----------------------------------------

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   rm(z0, nil) -> nil
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
S tuples:
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
K tuples:
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
Defined Rule Symbols:   eq_2, rm_2, ifrm_3

Defined Pair Symbols:   EQ_2, RM_2, IFRM_3, PURGE_1

Compound Symbols:   c3_1, c5_2, c6_1, c7_1, c9_2


----------------------------------------

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
We considered the (Usable) Rules:
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
   rm(z0, nil) -> nil
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
And the Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(EQ(x_1, x_2)) = 0
   POL(IFRM(x_1, x_2, x_3)) = [2]x_3
   POL(PURGE(x_1)) = [2]x_1^2
   POL(RM(x_1, x_2)) = [2]x_2
   POL(add(x_1, x_2)) = [1] + x_2
   POL(c3(x_1)) = x_1
   POL(c5(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1)) = x_1
   POL(c9(x_1, x_2)) = x_1 + x_2
   POL(eq(x_1, x_2)) = 0
   POL(false) = 0
   POL(ifrm(x_1, x_2, x_3)) = x_3
   POL(nil) = 0
   POL(rm(x_1, x_2)) = x_2
   POL(s(x_1)) = 0
   POL(true) = 0

----------------------------------------

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   eq(0, 0) -> true
   eq(0, s(z0)) -> false
   eq(s(z0), 0) -> false
   eq(s(z0), s(z1)) -> eq(z0, z1)
   rm(z0, nil) -> nil
   rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2))
   ifrm(true, z0, add(z1, z2)) -> rm(z0, z2)
   ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2))
Tuples:
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
S tuples:
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
K tuples:
   PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1))
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
Defined Rule Symbols:   eq_2, rm_2, ifrm_3

Defined Pair Symbols:   EQ_2, RM_2, IFRM_3, PURGE_1

Compound Symbols:   c3_1, c5_2, c6_1, c7_1, c9_2


----------------------------------------

(15) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1))
   IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2))
   IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2))
   EQ(s(z0), s(z1)) -> c3(EQ(z0, z1))
Now S is empty
----------------------------------------

(16)
BOUNDS(1, 1)

----------------------------------------

(17) RenamingProof (BOTH BOUNDS(ID, ID))
Renamed function symbols to avoid clashes with predefined symbol.
----------------------------------------

(18)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   eq(0', 0') -> true
   eq(0', s(X)) -> false
   eq(s(X), 0') -> false
   eq(s(X), s(Y)) -> eq(X, Y)
   rm(N, nil) -> nil
   rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
   ifrm(true, N, add(M, X)) -> rm(N, X)
   ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
   purge(nil) -> nil
   purge(add(N, X)) -> add(N, purge(rm(N, X)))

S is empty.
Rewrite Strategy: INNERMOST
----------------------------------------

(19) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
----------------------------------------

(20)
Obligation:
Innermost TRS:
Rules:
eq(0', 0') -> true
eq(0', s(X)) -> false
eq(s(X), 0') -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Types:
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
true :: true:false
s :: 0':s -> 0':s
false :: true:false
rm :: 0':s -> nil:add -> nil:add
nil :: nil:add
add :: 0':s -> nil:add -> nil:add
ifrm :: true:false -> 0':s -> nil:add -> nil:add
purge :: nil:add -> nil:add
hole_true:false1_0 :: true:false
hole_0':s2_0 :: 0':s
hole_nil:add3_0 :: nil:add
gen_0':s4_0 :: Nat -> 0':s
gen_nil:add5_0 :: Nat -> nil:add

----------------------------------------

(21) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
eq, rm, purge

They will be analysed ascendingly in the following order:
eq < rm
rm < purge

----------------------------------------

(22)
Obligation:
Innermost TRS:
Rules:
eq(0', 0') -> true
eq(0', s(X)) -> false
eq(s(X), 0') -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Types:
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
true :: true:false
s :: 0':s -> 0':s
false :: true:false
rm :: 0':s -> nil:add -> nil:add
nil :: nil:add
add :: 0':s -> nil:add -> nil:add
ifrm :: true:false -> 0':s -> nil:add -> nil:add
purge :: nil:add -> nil:add
hole_true:false1_0 :: true:false
hole_0':s2_0 :: 0':s
hole_nil:add3_0 :: nil:add
gen_0':s4_0 :: Nat -> 0':s
gen_nil:add5_0 :: Nat -> nil:add


Generator Equations:
gen_0':s4_0(0) <=> 0'
gen_0':s4_0(+(x, 1)) <=> s(gen_0':s4_0(x))
gen_nil:add5_0(0) <=> nil
gen_nil:add5_0(+(x, 1)) <=> add(0', gen_nil:add5_0(x))


The following defined symbols remain to be analysed:
eq, rm, purge

They will be analysed ascendingly in the following order:
eq < rm
rm < purge

----------------------------------------

(23) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
eq(gen_0':s4_0(n7_0), gen_0':s4_0(n7_0)) -> true, rt in Omega(1 + n7_0)

Induction Base:
eq(gen_0':s4_0(0), gen_0':s4_0(0)) ->_R^Omega(1)
true

Induction Step:
eq(gen_0':s4_0(+(n7_0, 1)), gen_0':s4_0(+(n7_0, 1))) ->_R^Omega(1)
eq(gen_0':s4_0(n7_0), gen_0':s4_0(n7_0)) ->_IH
true

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(24)
Complex Obligation (BEST)

----------------------------------------

(25)
Obligation:
Proved the lower bound n^1 for the following obligation:

Innermost TRS:
Rules:
eq(0', 0') -> true
eq(0', s(X)) -> false
eq(s(X), 0') -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Types:
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
true :: true:false
s :: 0':s -> 0':s
false :: true:false
rm :: 0':s -> nil:add -> nil:add
nil :: nil:add
add :: 0':s -> nil:add -> nil:add
ifrm :: true:false -> 0':s -> nil:add -> nil:add
purge :: nil:add -> nil:add
hole_true:false1_0 :: true:false
hole_0':s2_0 :: 0':s
hole_nil:add3_0 :: nil:add
gen_0':s4_0 :: Nat -> 0':s
gen_nil:add5_0 :: Nat -> nil:add


Generator Equations:
gen_0':s4_0(0) <=> 0'
gen_0':s4_0(+(x, 1)) <=> s(gen_0':s4_0(x))
gen_nil:add5_0(0) <=> nil
gen_nil:add5_0(+(x, 1)) <=> add(0', gen_nil:add5_0(x))


The following defined symbols remain to be analysed:
eq, rm, purge

They will be analysed ascendingly in the following order:
eq < rm
rm < purge

----------------------------------------

(26) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
----------------------------------------

(27)
BOUNDS(n^1, INF)

----------------------------------------

(28)
Obligation:
Innermost TRS:
Rules:
eq(0', 0') -> true
eq(0', s(X)) -> false
eq(s(X), 0') -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Types:
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
true :: true:false
s :: 0':s -> 0':s
false :: true:false
rm :: 0':s -> nil:add -> nil:add
nil :: nil:add
add :: 0':s -> nil:add -> nil:add
ifrm :: true:false -> 0':s -> nil:add -> nil:add
purge :: nil:add -> nil:add
hole_true:false1_0 :: true:false
hole_0':s2_0 :: 0':s
hole_nil:add3_0 :: nil:add
gen_0':s4_0 :: Nat -> 0':s
gen_nil:add5_0 :: Nat -> nil:add


Lemmas:
eq(gen_0':s4_0(n7_0), gen_0':s4_0(n7_0)) -> true, rt in Omega(1 + n7_0)


Generator Equations:
gen_0':s4_0(0) <=> 0'
gen_0':s4_0(+(x, 1)) <=> s(gen_0':s4_0(x))
gen_nil:add5_0(0) <=> nil
gen_nil:add5_0(+(x, 1)) <=> add(0', gen_nil:add5_0(x))


The following defined symbols remain to be analysed:
rm, purge

They will be analysed ascendingly in the following order:
rm < purge

----------------------------------------

(29) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
rm(gen_0':s4_0(0), gen_nil:add5_0(n470_0)) -> gen_nil:add5_0(0), rt in Omega(1 + n470_0)

Induction Base:
rm(gen_0':s4_0(0), gen_nil:add5_0(0)) ->_R^Omega(1)
nil

Induction Step:
rm(gen_0':s4_0(0), gen_nil:add5_0(+(n470_0, 1))) ->_R^Omega(1)
ifrm(eq(gen_0':s4_0(0), 0'), gen_0':s4_0(0), add(0', gen_nil:add5_0(n470_0))) ->_L^Omega(1)
ifrm(true, gen_0':s4_0(0), add(0', gen_nil:add5_0(n470_0))) ->_R^Omega(1)
rm(gen_0':s4_0(0), gen_nil:add5_0(n470_0)) ->_IH
gen_nil:add5_0(0)

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(30)
Obligation:
Innermost TRS:
Rules:
eq(0', 0') -> true
eq(0', s(X)) -> false
eq(s(X), 0') -> false
eq(s(X), s(Y)) -> eq(X, Y)
rm(N, nil) -> nil
rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) -> rm(N, X)
ifrm(false, N, add(M, X)) -> add(M, rm(N, X))
purge(nil) -> nil
purge(add(N, X)) -> add(N, purge(rm(N, X)))

Types:
eq :: 0':s -> 0':s -> true:false
0' :: 0':s
true :: true:false
s :: 0':s -> 0':s
false :: true:false
rm :: 0':s -> nil:add -> nil:add
nil :: nil:add
add :: 0':s -> nil:add -> nil:add
ifrm :: true:false -> 0':s -> nil:add -> nil:add
purge :: nil:add -> nil:add
hole_true:false1_0 :: true:false
hole_0':s2_0 :: 0':s
hole_nil:add3_0 :: nil:add
gen_0':s4_0 :: Nat -> 0':s
gen_nil:add5_0 :: Nat -> nil:add


Lemmas:
eq(gen_0':s4_0(n7_0), gen_0':s4_0(n7_0)) -> true, rt in Omega(1 + n7_0)
rm(gen_0':s4_0(0), gen_nil:add5_0(n470_0)) -> gen_nil:add5_0(0), rt in Omega(1 + n470_0)


Generator Equations:
gen_0':s4_0(0) <=> 0'
gen_0':s4_0(+(x, 1)) <=> s(gen_0':s4_0(x))
gen_nil:add5_0(0) <=> nil
gen_nil:add5_0(+(x, 1)) <=> add(0', gen_nil:add5_0(x))


The following defined symbols remain to be analysed:
purge