WORST_CASE(?,O(n^2))
proof of input_sv2JCKf2Px.trs
# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished


The Runtime Complexity (parallel-innermost) of the given CpxTRS could be proven to be BOUNDS(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)), 138 ms]
(8) CdtProblem
(9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 136 ms]
(10) CdtProblem
(11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 103 ms]
(12) CdtProblem
(13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 114 ms]
(14) CdtProblem
(15) CdtKnowledgeProof [FINISHED, 0 ms]
(16) BOUNDS(1, 1)


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

(0)
Obligation:
The Runtime Complexity (parallel-innermost) of the given CpxTRS could be proven to be BOUNDS(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: PARALLEL_INNERMOST
----------------------------------------

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST 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, s(z0)) -> c1
   EQ(s(z0), 0) -> c2
   PURGE(nil) -> c8
   RM(z0, nil) -> c4
   EQ(0, 0) -> c

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

(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


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(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
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(16)
BOUNDS(1, 1)