YES We show the termination of the TRS R: plus(x,|0|()) -> x plus(|0|(),y) -> y plus(s(x),y) -> s(plus(x,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: times#(s(x),y) -> plus#(y,times(x,y)) p3: times#(s(x),y) -> times#(x,y) p4: div#(x,y) -> quot#(x,y,y) p5: quot#(s(x),s(y),z) -> quot#(x,y,z) p6: quot#(x,|0|(),s(z)) -> div#(x,s(z)) p7: div#(div(x,y),z) -> div#(x,times(y,z)) p8: div#(div(x,y),z) -> times#(y,z) p9: eq#(s(x),s(y)) -> eq#(x,y) p10: divides#(y,x) -> eq#(x,times(div(x,y),y)) p11: divides#(y,x) -> times#(div(x,y),y) p12: divides#(y,x) -> div#(x,y) p13: prime#(s(s(x))) -> pr#(s(s(x)),s(x)) p14: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y)) p15: pr#(x,s(s(y))) -> divides#(s(s(y)),x) p16: if#(false(),x,y) -> pr#(x,y) and R consists of: r1: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: if(false(),x,y) -> pr(x,y) The estimated dependency graph contains the following SCCs: {p14, p16} {p4, p5, p6, p7} {p3} {p1} {p9} -- 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: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r4 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{28, x3 + 10} false_A = 20 pr#_A(x1,x2) = x2 + 2 s_A(x1) = max{17, x1 + 9} divides_A(x1,x2) = max{25, x1 - 4} times_A(x1,x2) = max{26, x1 - 22, x2 + 9} |0|_A = 21 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() = 13 pr#_A(x1,x2) = x2 s_A(x1) = x1 + 3 divides_A(x1,x2) = 1 times_A(x1,x2) = x1 + 2 |0|_A() = 9 if#!_A(x1,x2,x3) = x2 + 5 false!_A() = 5 pr#!_A(x1,x2) = x1 + x2 s!_A(x1) = x1 + 7 divides!_A(x1,x2) = 6 times!_A(x1,x2) = x1 + x2 + 8 |0|!_A() = 12 partial status: pi(if#) = [] pi(false) = [] pi(pr#) = [2] pi(s) = [1] pi(divides) = [] pi(times) = [2] pi(|0|) = [] 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: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r1, r2, r3, r4, r7 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: div#_A(x1,x2) = x1 div_A(x1,x2) = max{x1 + 6, x2 - 2} times_A(x1,x2) = max{15, x1 + 7, x2 + 1} quot#_A(x1,x2,x3) = x1 |0|_A = 0 s_A(x1) = x1 plus_A(x1,x2) = max{16, x1 + 2, 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: div#_A(x1,x2) = 0 div_A(x1,x2) = x1 + x2 + 2 times_A(x1,x2) = x1 + 1 quot#_A(x1,x2,x3) = 0 |0|_A() = 5 s_A(x1) = x1 + 1 plus_A(x1,x2) = x1 + x2 + 2 div#!_A(x1,x2) = x1 + 3 div!_A(x1,x2) = x1 + x2 + 8 times!_A(x1,x2) = x1 + x2 + 11 quot#!_A(x1,x2,x3) = x1 + 3 |0|!_A() = 2 s!_A(x1) = x1 + 4 plus!_A(x1,x2) = x1 + x2 + 6 partial status: pi(div#) = [1] pi(div) = [1] pi(times) = [] pi(quot#) = [1] pi(|0|) = [] pi(s) = [1] pi(plus) = [] The following rules are removed: p1, p4 -- 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: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: 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: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r7 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: div#_A(x1,x2) = 1 quot#_A(x1,x2,x3) = 1 |0|_A = 18 s_A(x1) = max{16, x1 + 6} div_A(x1,x2) = max{9, x1 + 6, x2 - 2} 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) = x3 |0|_A() = 6 s_A(x1) = 4 div_A(x1,x2) = x1 + x2 + 1 div#!_A(x1,x2) = x2 + 1 quot#!_A(x1,x2,x3) = x2 |0|!_A() = 22 s!_A(x1) = 7 div!_A(x1,x2) = 18 partial status: pi(div#) = [] pi(quot#) = [] pi(|0|) = [] pi(s) = [1] pi(div) = [] 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: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r4, r7 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: times#_A(x1,x2) = max{x1 + 16, x2 + 19} s_A(x1) = max{17, x1 + 15} times_A(x1,x2) = max{31, x1 - 5, x2 + 15} |0|_A = 6 div_A(x1,x2) = x1 + 15 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 + 6 times_A(x1,x2) = x1 + 8 |0|_A() = 9 div_A(x1,x2) = x1 + x2 + 7 times#!_A(x1,x2) = x1 s!_A(x1) = 13 times!_A(x1,x2) = x2 + 10 |0|!_A() = 12 div!_A(x1,x2) = x1 + x2 + 3 partial status: pi(times#) = [1, 2] pi(s) = [1] pi(times) = [] pi(|0|) = [] pi(div) = [1] 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) and R consists of: r1: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r4, r7 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = x1 s_A(x1) = x1 times_A(x1,x2) = max{5, x1 - 6, x2 + 2} |0|_A = 0 div_A(x1,x2) = x1 + 2 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 s_A(x1) = x1 + 1 times_A(x1,x2) = x1 + 2 |0|_A() = 9 div_A(x1,x2) = x1 + x2 + 1 plus#!_A(x1,x2) = 0 s!_A(x1) = 3 times!_A(x1,x2) = 2 |0|!_A() = 8 div!_A(x1,x2) = x1 + x2 partial status: pi(plus#) = [1] pi(s) = [1] pi(times) = [] pi(|0|) = [] pi(div) = [1] 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: plus(x,|0|()) -> x r2: plus(|0|(),y) -> y r3: plus(s(x),y) -> s(plus(x,y)) r4: times(|0|(),y) -> |0|() r5: times(s(|0|()),y) -> y r6: times(s(x),y) -> plus(y,times(x,y)) r7: div(|0|(),y) -> |0|() r8: div(x,y) -> quot(x,y,y) r9: quot(|0|(),s(y),z) -> |0|() r10: quot(s(x),s(y),z) -> quot(x,y,z) r11: quot(x,|0|(),s(z)) -> s(div(x,s(z))) r12: div(div(x,y),z) -> div(x,times(y,z)) r13: eq(|0|(),|0|()) -> true() r14: eq(s(x),|0|()) -> false() r15: eq(|0|(),s(y)) -> false() r16: eq(s(x),s(y)) -> eq(x,y) r17: divides(y,x) -> eq(x,times(div(x,y),y)) r18: prime(s(s(x))) -> pr(s(s(x)),s(x)) r19: pr(x,s(|0|())) -> true() r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y)) r21: if(true(),x,y) -> false() r22: if(false(),x,y) -> pr(x,y) The set of usable rules consists of r4, r7 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: eq#_A(x1,x2) = max{6, x1 - 1} s_A(x1) = max{16, x1 + 5} times_A(x1,x2) = max{23, x1 - 2, x2 + 3} |0|_A = 8 div_A(x1,x2) = 15 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 + 2 times_A(x1,x2) = x1 + 3 |0|_A() = 12 div_A(x1,x2) = x1 + 4 eq#!_A(x1,x2) = x1 s!_A(x1) = x1 + 7 times!_A(x1,x2) = 11 |0|!_A() = 10 div!_A(x1,x2) = 9 partial status: pi(eq#) = [] pi(s) = [] pi(times) = [] pi(|0|) = [] pi(div) = [] The following rules are removed: p1 Then no dependency pair remains.