YES We show the termination of the TRS R: p(|0|()) -> |0|() p(s(x)) -> x plus(x,|0|()) -> x plus(|0|(),y) -> y plus(s(x),y) -> s(plus(x,y)) plus(s(x),y) -> s(plus(p(s(x)),y)) plus(x,s(y)) -> s(plus(x,p(s(y)))) times(|0|(),y) -> |0|() times(s(|0|()),y) -> y times(s(x),y) -> plus(y,times(x,y)) div(|0|(),y) -> |0|() div(x,y) -> quot(x,y,y) quot(|0|(),s(y),z) -> |0|() quot(s(x),s(y),z) -> quot(x,y,z) quot(x,|0|(),s(z)) -> s(div(x,s(z))) div(div(x,y),z) -> div(x,times(y,z)) eq(|0|(),|0|()) -> true() eq(s(x),|0|()) -> false() eq(|0|(),s(y)) -> false() eq(s(x),s(y)) -> eq(x,y) divides(y,x) -> eq(x,times(div(x,y),y)) prime(s(s(x))) -> pr(s(s(x)),s(x)) pr(x,s(|0|())) -> true() pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) if(true(),x,y) -> false() if(false(),x,y) -> pr(x,y) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(x,y) p2: plus#(s(x),y) -> plus#(p(s(x)),y) p3: plus#(s(x),y) -> p#(s(x)) p4: plus#(x,s(y)) -> plus#(x,p(s(y))) p5: plus#(x,s(y)) -> p#(s(y)) p6: times#(s(x),y) -> plus#(y,times(x,y)) p7: times#(s(x),y) -> times#(x,y) p8: div#(x,y) -> quot#(x,y,y) p9: quot#(s(x),s(y),z) -> quot#(x,y,z) p10: quot#(x,|0|(),s(z)) -> div#(x,s(z)) p11: div#(div(x,y),z) -> div#(x,times(y,z)) p12: div#(div(x,y),z) -> times#(y,z) p13: eq#(s(x),s(y)) -> eq#(x,y) p14: divides#(y,x) -> eq#(x,times(div(x,y),y)) p15: divides#(y,x) -> times#(div(x,y),y) p16: divides#(y,x) -> div#(x,y) p17: prime#(s(s(x))) -> pr#(s(s(x)),s(x)) p18: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y)) p19: pr#(x,s(s(y))) -> divides#(s(s(y)),x) p20: if#(false(),x,y) -> pr#(x,y) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The estimated dependency graph contains the following SCCs: {p18, p20} {p8, p9, p10, p11} {p7} {p1, p2, p4} {p13} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: if#(false(),x,y) -> pr#(x,y) p2: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y)) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r17, r18, r19, r20, r21 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: if#_A(x1,x2,x3) = max{x1 - 8, x3 + 6} false_A = 4 pr#_A(x1,x2) = max{5, x2 + 1} s_A(x1) = x1 + 12 divides_A(x1,x2) = max{7, x1 - 44} eq_A(x1,x2) = 6 |0|_A = 11 true_A = 5 times_A(x1,x2) = max{x1 - 15, x2} div_A(x1,x2) = max{14, x1 + 12, x2 - 28} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: if#_A(x1,x2,x3) = x3 false_A() = 1 pr#_A(x1,x2) = x2 s_A(x1) = x1 + 2 divides_A(x1,x2) = 6 eq_A(x1,x2) = 2 |0|_A() = 4 true_A() = 2 times_A(x1,x2) = 1 div_A(x1,x2) = 1 if#!_A(x1,x2,x3) = x1 + 5 false!_A() = 6 pr#!_A(x1,x2) = 7 s!_A(x1) = 12 divides!_A(x1,x2) = 8 eq!_A(x1,x2) = x1 + 13 |0|!_A() = 11 true!_A() = 7 times!_A(x1,x2) = 10 div!_A(x1,x2) = 9 partial status: pi(if#) = [] pi(false) = [] pi(pr#) = [2] pi(s) = [] pi(divides) = [] pi(eq) = [] pi(|0|) = [] pi(true) = [] pi(times) = [] pi(div) = [1] The following rules are removed: p1, p2 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: div#(div(x,y),z) -> div#(x,times(y,z)) p2: div#(x,y) -> quot#(x,y,y) p3: quot#(x,|0|(),s(z)) -> div#(x,s(z)) p4: quot#(s(x),s(y),z) -> quot#(x,y,z) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r3, r11 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: div#_A(x1,x2) = 0 div_A(x1,x2) = x2 + 18 times_A(x1,x2) = 0 quot#_A(x1,x2,x3) = 0 |0|_A = 2 s_A(x1) = 5 plus_A(x1,x2) = max{4, x1} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: div#_A(x1,x2) = 0 div_A(x1,x2) = x1 + x2 times_A(x1,x2) = x1 + 1 quot#_A(x1,x2,x3) = 0 |0|_A() = 2 s_A(x1) = x1 + 2 plus_A(x1,x2) = x1 + 4 div#!_A(x1,x2) = x1 + 4 div!_A(x1,x2) = x1 + 9 times!_A(x1,x2) = x1 + 2 quot#!_A(x1,x2,x3) = x1 + 4 |0|!_A() = 8 s!_A(x1) = x1 + 3 plus!_A(x1,x2) = x1 + 5 partial status: pi(div#) = [] pi(div) = [2] pi(times) = [] pi(quot#) = [] pi(|0|) = [] pi(s) = [] pi(plus) = [1] The following rules are removed: p4 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: div#(div(x,y),z) -> div#(x,times(y,z)) p2: div#(x,y) -> quot#(x,y,y) p3: quot#(x,|0|(),s(z)) -> div#(x,s(z)) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The estimated dependency graph contains the following SCCs: {p1, p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: div#(div(x,y),z) -> div#(x,times(y,z)) p2: div#(x,y) -> quot#(x,y,y) p3: quot#(x,|0|(),s(z)) -> div#(x,s(z)) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r3, r4, r9, r11 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: div#_A(x1,x2) = 2 div_A(x1,x2) = max{20, x2 - 16} times_A(x1,x2) = max{x1 - 28, x2 + 7} quot#_A(x1,x2,x3) = 2 |0|_A = 3 s_A(x1) = max{27, x1 + 7} plus_A(x1,x2) = max{22, x1 + 13, x2 + 8} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: div#_A(x1,x2) = 0 div_A(x1,x2) = x1 + 1 times_A(x1,x2) = x1 + x2 + 2 quot#_A(x1,x2,x3) = 0 |0|_A() = 3 s_A(x1) = x1 + 5 plus_A(x1,x2) = x1 + x2 + 6 div#!_A(x1,x2) = x1 div!_A(x1,x2) = x1 + x2 + 7 times!_A(x1,x2) = x1 + 7 quot#!_A(x1,x2,x3) = x1 |0|!_A() = 10 s!_A(x1) = 11 plus!_A(x1,x2) = x2 + 12 partial status: pi(div#) = [] pi(div) = [] pi(times) = [2] pi(quot#) = [] pi(|0|) = [] pi(s) = [] pi(plus) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: div#(x,y) -> quot#(x,y,y) p2: quot#(x,|0|(),s(z)) -> div#(x,s(z)) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: div#(x,y) -> quot#(x,y,y) p2: quot#(x,|0|(),s(z)) -> div#(x,s(z)) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r3, r5, r6, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: div#_A(x1,x2) = x2 + 3 quot#_A(x1,x2,x3) = max{2, x2 - 1} |0|_A = 19 s_A(x1) = 15 plus_A(x1,x2) = max{x1 + 43, x2 + 27} p_A(x1) = max{14, x1 - 16} times_A(x1,x2) = max{39, x1 - 3, x2 - 27} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: div#_A(x1,x2) = x2 quot#_A(x1,x2,x3) = x2 |0|_A() = 3 s_A(x1) = 3 plus_A(x1,x2) = x1 + 10 p_A(x1) = 2 times_A(x1,x2) = x1 + 6 div#!_A(x1,x2) = 0 quot#!_A(x1,x2,x3) = x1 + 3 |0|!_A() = 1 s!_A(x1) = 2 plus!_A(x1,x2) = x2 + 1 p!_A(x1) = 0 times!_A(x1,x2) = x1 + 3 partial status: pi(div#) = [2] pi(quot#) = [] pi(|0|) = [] pi(s) = [] pi(plus) = [1, 2] pi(p) = [] pi(times) = [] The following rules are removed: p1, p2 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: times#(s(x),y) -> times#(x,y) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r8, r11 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: times#_A(x1,x2) = max{26, x1 + 6} s_A(x1) = x1 + 25 times_A(x1,x2) = max{40, x1 - 47, x2 - 47} |0|_A = 2 div_A(x1,x2) = max{20, x1 - 70, x2 + 1} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: times#_A(x1,x2) = x1 s_A(x1) = x1 + 3 times_A(x1,x2) = x1 + 6 |0|_A() = 3 div_A(x1,x2) = x1 + 1 times#!_A(x1,x2) = 0 s!_A(x1) = x1 + 8 times!_A(x1,x2) = x1 + 9 |0|!_A() = 3 div!_A(x1,x2) = x2 + 4 partial status: pi(times#) = [] pi(s) = [1] pi(times) = [] pi(|0|) = [] pi(div) = [] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(x,y) p2: plus#(x,s(y)) -> plus#(x,p(s(y))) p3: plus#(s(x),y) -> plus#(p(s(x)),y) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r2, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = x2 + 6 s_A(x1) = max{22, x1 + 15} p_A(x1) = max{15, x1 - 7} times_A(x1,x2) = max{37, x1 - 39, x2 - 39} |0|_A = 19 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 2 p_A(x1) = x1 times_A(x1,x2) = x1 + x2 + 2 |0|_A() = 3 plus#!_A(x1,x2) = x1 + 4 s!_A(x1) = 5 p!_A(x1) = x1 + 8 times!_A(x1,x2) = x1 + 6 |0|!_A() = 8 partial status: pi(plus#) = [2] pi(s) = [] pi(p) = [] pi(times) = [] pi(|0|) = [] The following rules are removed: p1, p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(p(s(x)),y) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(s(x),y) -> plus#(p(s(x)),y) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r2, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{1, x1 - 10} s_A(x1) = x1 + 15 p_A(x1) = max{14, x1 - 6} times_A(x1,x2) = max{72, x1 - 90, x2 - 90} |0|_A = 71 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus#_A(x1,x2) = 0 s_A(x1) = x1 + 2 p_A(x1) = x1 + 2 times_A(x1,x2) = x1 + 2 |0|_A() = 5 plus#!_A(x1,x2) = 4 s!_A(x1) = x1 + 5 p!_A(x1) = 5 times!_A(x1,x2) = x1 + 10 |0|!_A() = 10 partial status: pi(plus#) = [] pi(s) = [1] pi(p) = [] pi(times) = [] pi(|0|) = [] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: eq#(s(x),s(y)) -> eq#(x,y) and R consists of: r1: p(|0|()) -> |0|() r2: p(s(x)) -> x r3: plus(x,|0|()) -> x r4: plus(|0|(),y) -> y r5: plus(s(x),y) -> s(plus(x,y)) r6: plus(s(x),y) -> s(plus(p(s(x)),y)) r7: plus(x,s(y)) -> s(plus(x,p(s(y)))) r8: times(|0|(),y) -> |0|() r9: times(s(|0|()),y) -> y r10: times(s(x),y) -> plus(y,times(x,y)) r11: div(|0|(),y) -> |0|() r12: div(x,y) -> quot(x,y,y) r13: quot(|0|(),s(y),z) -> |0|() r14: quot(s(x),s(y),z) -> quot(x,y,z) r15: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r16: div(div(x,y),z) -> div(x,times(y,z)) r17: eq(|0|(),|0|()) -> true() r18: eq(s(x),|0|()) -> false() r19: eq(|0|(),s(y)) -> false() r20: eq(s(x),s(y)) -> eq(x,y) r21: divides(y,x) -> eq(x,times(div(x,y),y)) r22: prime(s(s(x))) -> pr(s(s(x)),s(x)) r23: pr(x,s(|0|())) -> true() r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r25: if(true(),x,y) -> false() r26: if(false(),x,y) -> pr(x,y) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: eq#_A(x1,x2) = max{x1 + 10, x2 + 11} s_A(x1) = x1 + 10 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: eq#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 8 eq#!_A(x1,x2) = x1 + x2 + 18 s!_A(x1) = 17 partial status: pi(eq#) = [1, 2] pi(s) = [] The following rules are removed: p1 Then no dependency pair remains.