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 r2, r4, r7, r9, r13, r14, r15, r16, r17, r21 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: if#_A(x1,x2,x3) = x1 + x3 + 2 false_A() = 2 pr#_A(x1,x2) = x2 + 3 s_A(x1) = x1 + 8 divides_A(x1,x2) = 5 plus_A(x1,x2) = x2 + 1 |0|_A() = 4 times_A(x1,x2) = 6 div_A(x1,x2) = 7 quot_A(x1,x2,x3) = x1 + x3 + 9 eq_A(x1,x2) = 4 true_A() = 3 if_A(x1,x2,x3) = x1 + 12 base order 2: precedence: false = quot = eq > |0| > s = times = true = if > pr# = plus = div > if# = divides partial status: pi(if#) = [] pi(false) = [] pi(pr#) = [2] pi(s) = [1] pi(divides) = [] pi(plus) = [] pi(|0|) = [] pi(times) = [] pi(div) = [] pi(quot) = [3] pi(eq) = [] pi(true) = [] pi(if) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: 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: (no SCCs) -- 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 r7, r9, r13, r14, r15, r16, r21 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: div#_A(x1,x2) = x1 + 5 div_A(x1,x2) = x1 + 7 times_A(x1,x2) = x2 quot#_A(x1,x2,x3) = x1 + 5 |0|_A() = 14 s_A(x1) = x1 + 6 quot_A(x1,x2,x3) = 15 eq_A(x1,x2) = 4 true_A() = 3 false_A() = 2 if_A(x1,x2,x3) = x3 + 7 base order 2: precedence: div > times = quot > |0| = true > false > if > s > div# = quot# > eq partial status: pi(div#) = [1] pi(div) = [] pi(times) = [] pi(quot#) = [1] pi(|0|) = [] pi(s) = [] pi(quot) = [] pi(eq) = [] pi(true) = [] pi(false) = [] pi(if) = [3] 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: 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, 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: 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, r4, r7, r9, r13, r14, r15, r16, r19, r21 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: div#_A(x1,x2) = x1 + 5 div_A(x1,x2) = x1 + 6 times_A(x1,x2) = 5 quot#_A(x1,x2,x3) = x1 + 5 |0|_A() = 4 s_A(x1) = x1 + 7 plus_A(x1,x2) = x1 + 6 quot_A(x1,x2,x3) = x1 + x3 + 8 eq_A(x1,x2) = 4 true_A() = 1 false_A() = 3 pr_A(x1,x2) = 2 if_A(x1,x2,x3) = x1 + 3 base order 2: precedence: if > plus = false > div# = quot# = s = pr > quot > div > times > |0| > true > eq partial status: pi(div#) = [] pi(div) = [] pi(times) = [] pi(quot#) = [] pi(|0|) = [] pi(s) = [] pi(plus) = [1] pi(quot) = [3] pi(eq) = [] pi(true) = [] pi(false) = [] pi(pr) = [] pi(if) = [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: 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 r3, r4, r5, r7, r11, r13, r14, r15, r16, r19, r21 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: 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) = 1 plus_A(x1,x2) = 2 times_A(x1,x2) = x2 + 8 div_A(x1,x2) = x1 + x2 + 9 quot_A(x1,x2,x3) = 2 eq_A(x1,x2) = 6 true_A() = 5 false_A() = 4 pr_A(x1,x2) = 8 if_A(x1,x2,x3) = x2 + 9 base order 2: precedence: times = quot > div = true > |0| = if > false > plus = pr > s > div# = eq > quot# partial status: pi(div#) = [] pi(quot#) = [] pi(|0|) = [] pi(s) = [] pi(plus) = [] pi(times) = [] pi(div) = [2] pi(quot) = [] pi(eq) = [] pi(true) = [] pi(false) = [] pi(pr) = [] pi(if) = [2] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: 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: (no SCCs) -- 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 r2, r5, r7, r9, r13, r14, r15, r16, r18, r19, r20, r21, r22 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: times#_A(x1,x2) = x1 s_A(x1) = x1 + 5 plus_A(x1,x2) = x2 + 6 |0|_A() = 2 times_A(x1,x2) = x2 + 6 div_A(x1,x2) = x1 + x2 + 7 quot_A(x1,x2,x3) = x3 + 6 eq_A(x1,x2) = x1 + 2 true_A() = 1 false_A() = 3 prime_A(x1) = x1 + 1 pr_A(x1,x2) = 4 if_A(x1,x2,x3) = 4 divides_A(x1,x2) = 3 base order 2: precedence: plus = quot = false = prime > s = pr = if > times# = times = div = divides > |0| = eq > true partial status: pi(times#) = [1] pi(s) = [1] pi(plus) = [] pi(|0|) = [] pi(times) = [] pi(div) = [2] pi(quot) = [3] pi(eq) = [] pi(true) = [] pi(false) = [] pi(prime) = [1] pi(pr) = [] pi(if) = [] pi(divides) = [] 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 r2, r5, r7, r9, r13, r14, r15, r16, r18, r19, r20, r21, r22 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus#_A(x1,x2) = x1 s_A(x1) = x1 + 5 plus_A(x1,x2) = x2 + 6 |0|_A() = 2 times_A(x1,x2) = x2 + 6 div_A(x1,x2) = x1 + x2 + 7 quot_A(x1,x2,x3) = x3 + 6 eq_A(x1,x2) = x1 + 2 true_A() = 1 false_A() = 3 prime_A(x1) = x1 + 1 pr_A(x1,x2) = 4 if_A(x1,x2,x3) = 4 divides_A(x1,x2) = 3 base order 2: precedence: plus = quot = false = prime > s = pr = if > plus# = times = div = divides > |0| = eq > true partial status: pi(plus#) = [1] pi(s) = [1] pi(plus) = [] pi(|0|) = [] pi(times) = [] pi(div) = [2] pi(quot) = [3] pi(eq) = [] pi(true) = [] pi(false) = [] pi(prime) = [1] pi(pr) = [] pi(if) = [] pi(divides) = [] 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 r2, r4, r7, r9, r13, r14, r15, r16, r19, r21 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: eq#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 13 plus_A(x1,x2) = x2 + 1 |0|_A() = 4 times_A(x1,x2) = x1 + 11 div_A(x1,x2) = x1 + 5 quot_A(x1,x2,x3) = x1 + x2 + x3 + 14 eq_A(x1,x2) = x1 + x2 + 14 true_A() = 1 false_A() = 3 pr_A(x1,x2) = 2 if_A(x1,x2,x3) = x2 + 27 base order 2: precedence: eq > times = true = if > div = quot = false > eq# = s = plus = pr > |0| partial status: pi(eq#) = [1] pi(s) = [1] pi(plus) = [] pi(|0|) = [] pi(times) = [] pi(div) = [1] pi(quot) = [1, 3] pi(eq) = [1, 2] pi(true) = [] pi(false) = [] pi(pr) = [] pi(if) = [2] The following rules are removed: p1 Then no dependency pair remains.