(0) Obligation:
Clauses:
p(0, 0).
p(s(X), X).
le(0, Y, true).
le(s(X), 0, false).
le(s(X), s(Y), B) :- le(X, Y, B).
minus(X, Y, Z) :- ','(le(X, Y, B), if(B, X, Y, Z)).
if(true, X, Y, 0).
if(false, X, Y, s(Z)) :- ','(p(X, X1), minus(X1, Y, Z)).
Query: minus(g,a,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
minusA(s(X1)) :- ','(lecB(false), minusA(X1)).
minusC(0, s(X1)) :- ','(lecD(0, false), minusA(X1)).
minusC(s(X1), s(X2)) :- ','(lecD(s(X1), false), minusC(X1, X2)).
leE(s(X1), s(X2), X3) :- leE(X1, X2, X3).
minusF(s(X1), 0, s(X2)) :- minusC(X1, X2).
minusF(s(X1), s(X2), X3) :- leE(X1, X2, X4).
minusF(s(X1), s(X2), s(X3)) :- ','(lecE(X1, X2, false), minusF(X1, s(X2), X3)).
Clauses:
minuscA(0) :- lecB(true).
minuscA(s(X1)) :- ','(lecB(false), minuscA(X1)).
minuscC(X1, 0) :- lecD(X1, true).
minuscC(0, s(X1)) :- ','(lecD(0, false), minuscA(X1)).
minuscC(s(X1), s(X2)) :- ','(lecD(s(X1), false), minuscC(X1, X2)).
lecE(0, X1, true).
lecE(s(X1), 0, false).
lecE(s(X1), s(X2), X3) :- lecE(X1, X2, X3).
minuscF(0, X1, 0).
minuscF(s(X1), 0, s(X2)) :- minuscC(X1, X2).
minuscF(s(X1), s(X2), 0) :- lecE(X1, X2, true).
minuscF(s(X1), s(X2), s(X3)) :- ','(lecE(X1, X2, false), minuscF(X1, s(X2), X3)).
lecD(0, true).
lecD(s(X1), false).
lecB(true).
Afs:
minusF(x1, x2, x3) = minusF(x1)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
minusF_in: (b,f,f)
minusC_in: (b,f)
minusA_in: (f)
leE_in: (b,f,f)
lecE_in: (b,f,b)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
MINUSF_IN_GAA(s(X1), 0, s(X2)) → U8_GAA(X1, X2, minusC_in_ga(X1, X2))
MINUSF_IN_GAA(s(X1), 0, s(X2)) → MINUSC_IN_GA(X1, X2)
MINUSC_IN_GA(0, s(X1)) → U3_GA(X1, lecD_in_gg(0, false))
U3_GA(X1, lecD_out_gg(0, false)) → U4_GA(X1, minusA_in_a(X1))
U3_GA(X1, lecD_out_gg(0, false)) → MINUSA_IN_A(X1)
MINUSA_IN_A(s(X1)) → U1_A(X1, lecB_in_g(false))
U1_A(X1, lecB_out_g(false)) → U2_A(X1, minusA_in_a(X1))
U1_A(X1, lecB_out_g(false)) → MINUSA_IN_A(X1)
MINUSC_IN_GA(s(X1), s(X2)) → U5_GA(X1, X2, lecD_in_gg(s(X1), false))
U5_GA(X1, X2, lecD_out_gg(s(X1), false)) → U6_GA(X1, X2, minusC_in_ga(X1, X2))
U5_GA(X1, X2, lecD_out_gg(s(X1), false)) → MINUSC_IN_GA(X1, X2)
MINUSF_IN_GAA(s(X1), s(X2), X3) → U9_GAA(X1, X2, X3, leE_in_gaa(X1, X2, X4))
MINUSF_IN_GAA(s(X1), s(X2), X3) → LEE_IN_GAA(X1, X2, X4)
LEE_IN_GAA(s(X1), s(X2), X3) → U7_GAA(X1, X2, X3, leE_in_gaa(X1, X2, X3))
LEE_IN_GAA(s(X1), s(X2), X3) → LEE_IN_GAA(X1, X2, X3)
MINUSF_IN_GAA(s(X1), s(X2), s(X3)) → U10_GAA(X1, X2, X3, lecE_in_gag(X1, X2, false))
U10_GAA(X1, X2, X3, lecE_out_gag(X1, X2, false)) → U11_GAA(X1, X2, X3, minusF_in_gaa(X1, s(X2), X3))
U10_GAA(X1, X2, X3, lecE_out_gag(X1, X2, false)) → MINUSF_IN_GAA(X1, s(X2), X3)
The TRS R consists of the following rules:
lecD_in_gg(0, true) → lecD_out_gg(0, true)
lecD_in_gg(s(X1), false) → lecD_out_gg(s(X1), false)
lecB_in_g(true) → lecB_out_g(true)
lecE_in_gag(0, X1, true) → lecE_out_gag(0, X1, true)
lecE_in_gag(s(X1), 0, false) → lecE_out_gag(s(X1), 0, false)
lecE_in_gag(s(X1), s(X2), X3) → U21_gag(X1, X2, X3, lecE_in_gag(X1, X2, X3))
U21_gag(X1, X2, X3, lecE_out_gag(X1, X2, X3)) → lecE_out_gag(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
minusF_in_gaa(
x1,
x2,
x3) =
minusF_in_gaa(
x1)
s(
x1) =
s(
x1)
minusC_in_ga(
x1,
x2) =
minusC_in_ga(
x1)
0 =
0
lecD_in_gg(
x1,
x2) =
lecD_in_gg(
x1,
x2)
true =
true
lecD_out_gg(
x1,
x2) =
lecD_out_gg(
x1,
x2)
false =
false
minusA_in_a(
x1) =
minusA_in_a
lecB_in_g(
x1) =
lecB_in_g(
x1)
lecB_out_g(
x1) =
lecB_out_g(
x1)
leE_in_gaa(
x1,
x2,
x3) =
leE_in_gaa(
x1)
lecE_in_gag(
x1,
x2,
x3) =
lecE_in_gag(
x1,
x3)
lecE_out_gag(
x1,
x2,
x3) =
lecE_out_gag(
x1,
x3)
U21_gag(
x1,
x2,
x3,
x4) =
U21_gag(
x1,
x3,
x4)
MINUSF_IN_GAA(
x1,
x2,
x3) =
MINUSF_IN_GAA(
x1)
U8_GAA(
x1,
x2,
x3) =
U8_GAA(
x1,
x3)
MINUSC_IN_GA(
x1,
x2) =
MINUSC_IN_GA(
x1)
U3_GA(
x1,
x2) =
U3_GA(
x2)
U4_GA(
x1,
x2) =
U4_GA(
x2)
MINUSA_IN_A(
x1) =
MINUSA_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
U2_A(
x1,
x2) =
U2_A(
x2)
U5_GA(
x1,
x2,
x3) =
U5_GA(
x1,
x3)
U6_GA(
x1,
x2,
x3) =
U6_GA(
x1,
x3)
U9_GAA(
x1,
x2,
x3,
x4) =
U9_GAA(
x1,
x4)
LEE_IN_GAA(
x1,
x2,
x3) =
LEE_IN_GAA(
x1)
U7_GAA(
x1,
x2,
x3,
x4) =
U7_GAA(
x1,
x4)
U10_GAA(
x1,
x2,
x3,
x4) =
U10_GAA(
x1,
x4)
U11_GAA(
x1,
x2,
x3,
x4) =
U11_GAA(
x1,
x4)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MINUSF_IN_GAA(s(X1), 0, s(X2)) → U8_GAA(X1, X2, minusC_in_ga(X1, X2))
MINUSF_IN_GAA(s(X1), 0, s(X2)) → MINUSC_IN_GA(X1, X2)
MINUSC_IN_GA(0, s(X1)) → U3_GA(X1, lecD_in_gg(0, false))
U3_GA(X1, lecD_out_gg(0, false)) → U4_GA(X1, minusA_in_a(X1))
U3_GA(X1, lecD_out_gg(0, false)) → MINUSA_IN_A(X1)
MINUSA_IN_A(s(X1)) → U1_A(X1, lecB_in_g(false))
U1_A(X1, lecB_out_g(false)) → U2_A(X1, minusA_in_a(X1))
U1_A(X1, lecB_out_g(false)) → MINUSA_IN_A(X1)
MINUSC_IN_GA(s(X1), s(X2)) → U5_GA(X1, X2, lecD_in_gg(s(X1), false))
U5_GA(X1, X2, lecD_out_gg(s(X1), false)) → U6_GA(X1, X2, minusC_in_ga(X1, X2))
U5_GA(X1, X2, lecD_out_gg(s(X1), false)) → MINUSC_IN_GA(X1, X2)
MINUSF_IN_GAA(s(X1), s(X2), X3) → U9_GAA(X1, X2, X3, leE_in_gaa(X1, X2, X4))
MINUSF_IN_GAA(s(X1), s(X2), X3) → LEE_IN_GAA(X1, X2, X4)
LEE_IN_GAA(s(X1), s(X2), X3) → U7_GAA(X1, X2, X3, leE_in_gaa(X1, X2, X3))
LEE_IN_GAA(s(X1), s(X2), X3) → LEE_IN_GAA(X1, X2, X3)
MINUSF_IN_GAA(s(X1), s(X2), s(X3)) → U10_GAA(X1, X2, X3, lecE_in_gag(X1, X2, false))
U10_GAA(X1, X2, X3, lecE_out_gag(X1, X2, false)) → U11_GAA(X1, X2, X3, minusF_in_gaa(X1, s(X2), X3))
U10_GAA(X1, X2, X3, lecE_out_gag(X1, X2, false)) → MINUSF_IN_GAA(X1, s(X2), X3)
The TRS R consists of the following rules:
lecD_in_gg(0, true) → lecD_out_gg(0, true)
lecD_in_gg(s(X1), false) → lecD_out_gg(s(X1), false)
lecB_in_g(true) → lecB_out_g(true)
lecE_in_gag(0, X1, true) → lecE_out_gag(0, X1, true)
lecE_in_gag(s(X1), 0, false) → lecE_out_gag(s(X1), 0, false)
lecE_in_gag(s(X1), s(X2), X3) → U21_gag(X1, X2, X3, lecE_in_gag(X1, X2, X3))
U21_gag(X1, X2, X3, lecE_out_gag(X1, X2, X3)) → lecE_out_gag(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
minusF_in_gaa(
x1,
x2,
x3) =
minusF_in_gaa(
x1)
s(
x1) =
s(
x1)
minusC_in_ga(
x1,
x2) =
minusC_in_ga(
x1)
0 =
0
lecD_in_gg(
x1,
x2) =
lecD_in_gg(
x1,
x2)
true =
true
lecD_out_gg(
x1,
x2) =
lecD_out_gg(
x1,
x2)
false =
false
minusA_in_a(
x1) =
minusA_in_a
lecB_in_g(
x1) =
lecB_in_g(
x1)
lecB_out_g(
x1) =
lecB_out_g(
x1)
leE_in_gaa(
x1,
x2,
x3) =
leE_in_gaa(
x1)
lecE_in_gag(
x1,
x2,
x3) =
lecE_in_gag(
x1,
x3)
lecE_out_gag(
x1,
x2,
x3) =
lecE_out_gag(
x1,
x3)
U21_gag(
x1,
x2,
x3,
x4) =
U21_gag(
x1,
x3,
x4)
MINUSF_IN_GAA(
x1,
x2,
x3) =
MINUSF_IN_GAA(
x1)
U8_GAA(
x1,
x2,
x3) =
U8_GAA(
x1,
x3)
MINUSC_IN_GA(
x1,
x2) =
MINUSC_IN_GA(
x1)
U3_GA(
x1,
x2) =
U3_GA(
x2)
U4_GA(
x1,
x2) =
U4_GA(
x2)
MINUSA_IN_A(
x1) =
MINUSA_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
U2_A(
x1,
x2) =
U2_A(
x2)
U5_GA(
x1,
x2,
x3) =
U5_GA(
x1,
x3)
U6_GA(
x1,
x2,
x3) =
U6_GA(
x1,
x3)
U9_GAA(
x1,
x2,
x3,
x4) =
U9_GAA(
x1,
x4)
LEE_IN_GAA(
x1,
x2,
x3) =
LEE_IN_GAA(
x1)
U7_GAA(
x1,
x2,
x3,
x4) =
U7_GAA(
x1,
x4)
U10_GAA(
x1,
x2,
x3,
x4) =
U10_GAA(
x1,
x4)
U11_GAA(
x1,
x2,
x3,
x4) =
U11_GAA(
x1,
x4)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 13 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LEE_IN_GAA(s(X1), s(X2), X3) → LEE_IN_GAA(X1, X2, X3)
The TRS R consists of the following rules:
lecD_in_gg(0, true) → lecD_out_gg(0, true)
lecD_in_gg(s(X1), false) → lecD_out_gg(s(X1), false)
lecB_in_g(true) → lecB_out_g(true)
lecE_in_gag(0, X1, true) → lecE_out_gag(0, X1, true)
lecE_in_gag(s(X1), 0, false) → lecE_out_gag(s(X1), 0, false)
lecE_in_gag(s(X1), s(X2), X3) → U21_gag(X1, X2, X3, lecE_in_gag(X1, X2, X3))
U21_gag(X1, X2, X3, lecE_out_gag(X1, X2, X3)) → lecE_out_gag(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
0 =
0
lecD_in_gg(
x1,
x2) =
lecD_in_gg(
x1,
x2)
true =
true
lecD_out_gg(
x1,
x2) =
lecD_out_gg(
x1,
x2)
false =
false
lecB_in_g(
x1) =
lecB_in_g(
x1)
lecB_out_g(
x1) =
lecB_out_g(
x1)
lecE_in_gag(
x1,
x2,
x3) =
lecE_in_gag(
x1,
x3)
lecE_out_gag(
x1,
x2,
x3) =
lecE_out_gag(
x1,
x3)
U21_gag(
x1,
x2,
x3,
x4) =
U21_gag(
x1,
x3,
x4)
LEE_IN_GAA(
x1,
x2,
x3) =
LEE_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(8) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(9) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LEE_IN_GAA(s(X1), s(X2), X3) → LEE_IN_GAA(X1, X2, X3)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
LEE_IN_GAA(
x1,
x2,
x3) =
LEE_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(10) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LEE_IN_GAA(s(X1)) → LEE_IN_GAA(X1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- LEE_IN_GAA(s(X1)) → LEE_IN_GAA(X1)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MINUSF_IN_GAA(s(X1), s(X2), s(X3)) → U10_GAA(X1, X2, X3, lecE_in_gag(X1, X2, false))
U10_GAA(X1, X2, X3, lecE_out_gag(X1, X2, false)) → MINUSF_IN_GAA(X1, s(X2), X3)
The TRS R consists of the following rules:
lecD_in_gg(0, true) → lecD_out_gg(0, true)
lecD_in_gg(s(X1), false) → lecD_out_gg(s(X1), false)
lecB_in_g(true) → lecB_out_g(true)
lecE_in_gag(0, X1, true) → lecE_out_gag(0, X1, true)
lecE_in_gag(s(X1), 0, false) → lecE_out_gag(s(X1), 0, false)
lecE_in_gag(s(X1), s(X2), X3) → U21_gag(X1, X2, X3, lecE_in_gag(X1, X2, X3))
U21_gag(X1, X2, X3, lecE_out_gag(X1, X2, X3)) → lecE_out_gag(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
0 =
0
lecD_in_gg(
x1,
x2) =
lecD_in_gg(
x1,
x2)
true =
true
lecD_out_gg(
x1,
x2) =
lecD_out_gg(
x1,
x2)
false =
false
lecB_in_g(
x1) =
lecB_in_g(
x1)
lecB_out_g(
x1) =
lecB_out_g(
x1)
lecE_in_gag(
x1,
x2,
x3) =
lecE_in_gag(
x1,
x3)
lecE_out_gag(
x1,
x2,
x3) =
lecE_out_gag(
x1,
x3)
U21_gag(
x1,
x2,
x3,
x4) =
U21_gag(
x1,
x3,
x4)
MINUSF_IN_GAA(
x1,
x2,
x3) =
MINUSF_IN_GAA(
x1)
U10_GAA(
x1,
x2,
x3,
x4) =
U10_GAA(
x1,
x4)
We have to consider all (P,R,Pi)-chains
(15) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MINUSF_IN_GAA(s(X1), s(X2), s(X3)) → U10_GAA(X1, X2, X3, lecE_in_gag(X1, X2, false))
U10_GAA(X1, X2, X3, lecE_out_gag(X1, X2, false)) → MINUSF_IN_GAA(X1, s(X2), X3)
The TRS R consists of the following rules:
lecE_in_gag(s(X1), 0, false) → lecE_out_gag(s(X1), 0, false)
lecE_in_gag(s(X1), s(X2), X3) → U21_gag(X1, X2, X3, lecE_in_gag(X1, X2, X3))
U21_gag(X1, X2, X3, lecE_out_gag(X1, X2, X3)) → lecE_out_gag(s(X1), s(X2), X3)
lecE_in_gag(0, X1, true) → lecE_out_gag(0, X1, true)
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
0 =
0
true =
true
false =
false
lecE_in_gag(
x1,
x2,
x3) =
lecE_in_gag(
x1,
x3)
lecE_out_gag(
x1,
x2,
x3) =
lecE_out_gag(
x1,
x3)
U21_gag(
x1,
x2,
x3,
x4) =
U21_gag(
x1,
x3,
x4)
MINUSF_IN_GAA(
x1,
x2,
x3) =
MINUSF_IN_GAA(
x1)
U10_GAA(
x1,
x2,
x3,
x4) =
U10_GAA(
x1,
x4)
We have to consider all (P,R,Pi)-chains
(17) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUSF_IN_GAA(s(X1)) → U10_GAA(X1, lecE_in_gag(X1, false))
U10_GAA(X1, lecE_out_gag(X1, false)) → MINUSF_IN_GAA(X1)
The TRS R consists of the following rules:
lecE_in_gag(s(X1), false) → lecE_out_gag(s(X1), false)
lecE_in_gag(s(X1), X3) → U21_gag(X1, X3, lecE_in_gag(X1, X3))
U21_gag(X1, X3, lecE_out_gag(X1, X3)) → lecE_out_gag(s(X1), X3)
lecE_in_gag(0, true) → lecE_out_gag(0, true)
The set Q consists of the following terms:
lecE_in_gag(x0, x1)
U21_gag(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(19) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- U10_GAA(X1, lecE_out_gag(X1, false)) → MINUSF_IN_GAA(X1)
The graph contains the following edges 1 >= 1, 2 > 1
- MINUSF_IN_GAA(s(X1)) → U10_GAA(X1, lecE_in_gag(X1, false))
The graph contains the following edges 1 > 1
(20) YES
(21) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MINUSC_IN_GA(s(X1), s(X2)) → U5_GA(X1, X2, lecD_in_gg(s(X1), false))
U5_GA(X1, X2, lecD_out_gg(s(X1), false)) → MINUSC_IN_GA(X1, X2)
The TRS R consists of the following rules:
lecD_in_gg(0, true) → lecD_out_gg(0, true)
lecD_in_gg(s(X1), false) → lecD_out_gg(s(X1), false)
lecB_in_g(true) → lecB_out_g(true)
lecE_in_gag(0, X1, true) → lecE_out_gag(0, X1, true)
lecE_in_gag(s(X1), 0, false) → lecE_out_gag(s(X1), 0, false)
lecE_in_gag(s(X1), s(X2), X3) → U21_gag(X1, X2, X3, lecE_in_gag(X1, X2, X3))
U21_gag(X1, X2, X3, lecE_out_gag(X1, X2, X3)) → lecE_out_gag(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
0 =
0
lecD_in_gg(
x1,
x2) =
lecD_in_gg(
x1,
x2)
true =
true
lecD_out_gg(
x1,
x2) =
lecD_out_gg(
x1,
x2)
false =
false
lecB_in_g(
x1) =
lecB_in_g(
x1)
lecB_out_g(
x1) =
lecB_out_g(
x1)
lecE_in_gag(
x1,
x2,
x3) =
lecE_in_gag(
x1,
x3)
lecE_out_gag(
x1,
x2,
x3) =
lecE_out_gag(
x1,
x3)
U21_gag(
x1,
x2,
x3,
x4) =
U21_gag(
x1,
x3,
x4)
MINUSC_IN_GA(
x1,
x2) =
MINUSC_IN_GA(
x1)
U5_GA(
x1,
x2,
x3) =
U5_GA(
x1,
x3)
We have to consider all (P,R,Pi)-chains
(22) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(23) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MINUSC_IN_GA(s(X1), s(X2)) → U5_GA(X1, X2, lecD_in_gg(s(X1), false))
U5_GA(X1, X2, lecD_out_gg(s(X1), false)) → MINUSC_IN_GA(X1, X2)
The TRS R consists of the following rules:
lecD_in_gg(s(X1), false) → lecD_out_gg(s(X1), false)
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
lecD_in_gg(
x1,
x2) =
lecD_in_gg(
x1,
x2)
lecD_out_gg(
x1,
x2) =
lecD_out_gg(
x1,
x2)
false =
false
MINUSC_IN_GA(
x1,
x2) =
MINUSC_IN_GA(
x1)
U5_GA(
x1,
x2,
x3) =
U5_GA(
x1,
x3)
We have to consider all (P,R,Pi)-chains
(24) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUSC_IN_GA(s(X1)) → U5_GA(X1, lecD_in_gg(s(X1), false))
U5_GA(X1, lecD_out_gg(s(X1), false)) → MINUSC_IN_GA(X1)
The TRS R consists of the following rules:
lecD_in_gg(s(X1), false) → lecD_out_gg(s(X1), false)
The set Q consists of the following terms:
lecD_in_gg(x0, x1)
We have to consider all (P,Q,R)-chains.
(26) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- U5_GA(X1, lecD_out_gg(s(X1), false)) → MINUSC_IN_GA(X1)
The graph contains the following edges 1 >= 1, 2 > 1
- MINUSC_IN_GA(s(X1)) → U5_GA(X1, lecD_in_gg(s(X1), false))
The graph contains the following edges 1 > 1
(27) YES