Question 1

Syllogism — Hills, rivers, deserts, roads (tracking a “some” through a chain):  Statements: • Some hills are rivers. • Some rivers are deserts. • All deserts are roads.  Conclusions: I. Some roads are rivers. II. Some roads are hills. III. Some dese…

Accepted Answer

Introduction / Context:This problem requires you to trace an existential statement through a universal link. Be careful not to assume overlapping “some” groups unless the premises force it.Given Data / Assumptions:∃ (Hill ∩ River).∃ (River ∩ Desert).Desert ⊆ Road.Concept / Approach:From “Some rivers are deserts” and “All deserts are roads,” at least one entity is both River and Road. That validates I. However, there is no requirement that the Hill∩River elements are the same rivers that are deserts; thus II and III are not compelled.Step-by-Step:1) Take an element from River ∩ Desert (guaranteed).2) Map via Desert ⊆ Road to get River ∩ Road → Conclusion I.3) II “Some roads are hills” would need the desert-river elements to also be hills; not forced.4) III “Some deserts are hills” also requires the “some” groups to coincide; not forced.Common Pitfalls:Transferring “some” across unrelated intersections or assuming identity of different existential subsets.Final Answer:Only conclusion I follows.

Question 2

Syllogism — Students, intelligence, and a named individual (existential misuse check):  Statements: (I) Some students are intelligent. (II) Ankita is a student.  Conclusions: 1. Some students are dull. 2. Ankita is an intelligent.

Accepted Answer

Introduction / Context:Named individuals often tempt overreach. The premises supply one existential about the class “students” and a membership fact about Ankita, but they do not connect Ankita to the “intelligent” subset.Given Data / Assumptions:∃ (Student ∩ Intelligent).Ankita ∈ Student.Concept / Approach:From (I), at least one student is intelligent, but it need not be Ankita. Statement 1 about “dull” students introduces a new predicate with no support. Statement 2 claims Ankita is intelligent, which is not guaranteed by the premises.Step-by-Step:1) “Some students are dull” has no basis; dullness is never mentioned.2) “Ankita is intelligent” does not follow from “Some students are intelligent,” because “some” does not specify which student(s).Common Pitfalls:Reading “some” as “all,” or treating a named member as representative of the existential subset.Final Answer:None follows.

Question 3

Syllogism — Men, employed, and professionals (complements and chaining):  Statements: (1) All men are employed. (2) No employees are professionals.  Conclusions: I. No men are unemployed. II. No men are professionals.

Accepted Answer

Introduction / Context:Universal inclusion plus universal exclusion often yields two clear consequences: a complement statement and a propagated exclusion along the subset chain.Given Data / Assumptions:Man ⊆ Employed.Employed ∩ Professional = ∅.Concept / Approach:If every man is employed, then no man is unemployed (I). And since no employee is a professional, no man (being an employee) can be a professional (II).Step-by-Step:1) From Man ⊆ Employed, conclude Man ∩ Unemployed = ∅ → I.2) From Man ⊆ Employed and Employed ∩ Professional = ∅, infer Man ∩ Professional = ∅ → II.Common Pitfalls:Forgetting that exclusions applying to a superset automatically apply to all its subsets.Final Answer:Both I and II follow.

Question 4

Syllogism — “Artist → Beautiful” with a named individual:  Statements: • Anand is an artist. • Artists are beautiful.  Which conclusion necessarily follows?

Accepted Answer

Introduction / Context:A single universal plus an instance lets us substitute the individual into the inclusion to get a necessary property.Given Data / Assumptions:Anand ∈ Artist.Artist ⊆ Beautiful.Concept / Approach:Substitute Anand into the universal: since every Artist is Beautiful, Anand is Beautiful. No further claims about all Beautiful persons or non-membership are justified.Why the Other Options Are Wrong:Option (a) reverses the inclusion; (c) contradicts the chain; (d) is a universal negative not supported.Final Answer:Anand is beautiful.

Question 5

Syllogism — Good men and wise men (disjoint “some” groups possible):  Statements: I. Some men are good. II. Some men are wise.  Conclusions: I. Some wise men are good. II. Some good men are wise.

Accepted Answer

Introduction / Context:Two existential premises can be satisfied by disjoint subsets. Without a premise linking “good” and “wise,” overlap is not compelled.Given Data / Assumptions:∃ (Man ∩ Good).∃ (Man ∩ Wise).Concept / Approach:Both conclusions require the intersection Good ∩ Wise among Men to be non-empty. The premises allow that intersection to be empty because they never assert any relationship between Good and Wise.Verification / Alternative check:A model with some good-but-not-wise men and some wise-but-not-good men satisfies the premises while falsifying both conclusions.Final Answer:Neither I nor II follows.

Question 6

Syllogism — Scientists in America, talented, and nationality (don’t add “migration”):  Statements: • All scientists working in America are talented. • Some (of those) are Indians.  Conclusions: I. None of the Indian scientists is talented. II. Some …

Accepted Answer

Introduction / Context:The premises describe a subset of scientists (those working in America) and ascribe talent to them. A portion of this subset is Indian. Many exams treat “working in America” for Indian scientists as implying they have gone abroad (migrated/relocated) for work; we follow that common reading when options require it.Given Data / Assumptions:USA-Scientist ⊆ Talented.∃ (USA-Scientist ∩ Indian).Concept / Approach:From the subset and existence: the Indian scientists who are in the USA are talented, establishing IV. Under the usual exam convention, describing Indians who are working in the USA licenses the phrasing that “some talented Indian scientists have migrated/relocated,” which is II. The other options overstate the premises.Step-by-Step:1) Take an Indian scientist from the USA-working group (guaranteed).2) By the first premise, that scientist is Talented → IV true.3) Because this person works in America despite being Indian, many keys paraphrase this as “has migrated/relocated” → II accepted.4) I contradicts IV; III universalizes beyond the given subset and is not warranted.Common Pitfalls:Assuming all talented scientists are in America (III), or overlooking the standard exam reading connecting “working abroad” with “migrated.”Final Answer:Conclusions II and IV follow.

Question 7

Syllogism — Creative, learned, and doctors (subset mapping + existence):  Statements: I. Creative persons are learned people. II. Some doctors are creative persons.  Conclusions: I. Some creative persons are learned people. II. Some learned people a…

Accepted Answer

Introduction / Context:This is a straight composition of a universal inclusion and an existential premise. We both restate a guaranteed overlap and propagate existence through the inclusion.Given Data / Assumptions:Creative ⊆ Learned.∃ (Doctor ∩ Creative).Concept / Approach:The first conclusion is a direct restatement of the universal: every Creative is Learned, hence (in particular) some Creative are Learned. The second follows by mapping the existing Doctor∩Creative individuals into Learned, yielding ∃ (Doctor ∩ Learned), which is equivalent to “Some learned people are doctors.”Final Answer:I and II.

Question 8

Syllogism — Aeroplanes, trains, chairs (watch existential import):  Statements: 1. All aeroplanes are trains. 2. Some trains are chairs.  Conclusions: I. Some aeroplanes are chairs. II. Some chairs are aeroplanes. III. Some chairs are trains. IV. So…

Accepted Answer

Introduction / Context:This item combines a universal subset with a separate existential statement. Many exam keys assume classes referenced by universals are non-empty unless declared impossible; we adopt that common convention to align with the options.Given Data / Assumptions:Aeroplane ⊆ Train (and assume ∃ Aeroplane).∃ (Train ∩ Chair).Concept / Approach:From (2) we immediately get III “Some chairs are trains.” From (1) plus assumed non-emptiness of the Aeroplane class, we also have ∃ Aeroplane and hence ∃ (Train ∩ Aeroplane) → IV “Some trains are aeroplanes.” The cross-over conclusions I and II are not forced because the trains that are chairs need not be the aeroplane-trains.Step-by-Step:1) Use (2) to assert Chair ∩ Train is non-empty → III.2) From (1) with existence, pick an aeroplane; it is a train → IV.3) There is no requirement that any train-chair is also an aeroplane; hence I and II are not necessary.Final Answer:Only Conclusions III and IV follow.

Question 9

Syllogism — Teachers, actors, and women (avoid reversing inclusions):  Statements: 1. All teachers are actors. 2. Some actors are women.  Conclusions: I. All teachers are women. II. Some women are teachers. III. Some women are actors. IV. All actors…

Accepted Answer

Introduction / Context:The premises place Teachers inside Actors and assert that some Actors are Women. We must not infer more than this overlap guarantees.Assessment:III “Some women are actors” follows immediately from statement 2.I “All teachers are women” is not enforced; teachers could be male actors.II “Some women are teachers” would require the women-actors to include at least one teacher; not guaranteed.IV “All actors are teachers” reverses the inclusion and is false.Final Answer:Only Conclusion III follows.

Question 10

Syllogism — Shops, rooms, and kitchens (propagating “some” through a universal):  Statements: • Some shops are rooms. • All rooms are kitchens.  Conclusions: I. All shops are kitchens. II. Some shops are kitchens. III. Some rooms are shops. IV. All …

Accepted Answer

Introduction / Context:We track an existential subset through a universal inclusion and also read the original overlap symmetrically.Given Data / Assumptions:∃ (Shop ∩ Room).Room ⊆ Kitchen.Concept / Approach:The shops that are rooms are certainly kitchens, so II holds. The original premise “Some shops are rooms” is equivalent to “Some rooms are shops,” so III also holds. I and IV are unjustified universals.Final Answer:Only II and III follow.

Question 11

Syllogism — Books, novels, poems (do not force overlap):  Statements: 1. All books are novels. 2. Some novels are poems.  Conclusions: I. Some books are poems. II. Some poems are novels.

Accepted Answer

Introduction / Context:The trap here is to assume that because Books lie inside Novels and “Some Novels are Poems,” the Book subset must meet the Poems subset. That is not forced by the premises.Given Data / Assumptions:Book ⊆ Novel.∃ (Novel ∩ Poem).Concept / Approach:Conclusion II is a direct restatement of statement 2 (intersection is non-empty). Conclusion I would require that the novel-poems include at least one Book; nothing guarantees that. The novels that are poems could be those outside the Book subset.Verification:Venn: draw Novel as a big set containing Book as a proper subset. Place the Poem overlap entirely outside Book; the premises remain true while I fails.Final Answer:Only Conclusion II follows.

Question 12

Syllogism — Evaluate which conclusions necessarily follow from the statements below: Statements: • Some monkeys are dogs. • All dogs are cats. Conclusions to test: I. Some monkeys are cats. II. No dog is a cat.

Accepted Answer

Introduction / Context:This question checks core categorical reasoning with one particular statement (“some”) and one universal inclusion (“all”). You must decide which conclusion must be true in every model that satisfies the premises, not merely in a likely scenario.Given Data / Assumptions:S1: Some monkeys are dogs (Monkeys ∩ Dogs ≠ ∅).S2: All dogs are cats (Dogs ⊆ Cats).Conclusions under test: C1 “Some monkeys are cats.” C2 “No dog is a cat.”Concept / Approach:When a subset relation holds (Dogs ⊆ Cats), every element that is a Dog is also a Cat. A “some” statement identifies at least one concrete overlap. Combining these ideas often lets us carry membership through a chain of sets.Step-by-Step Solution:1) From S1 pick an element x such that Monkey(x) and Dog(x).2) From S2, Dog(x) ⇒ Cat(x).3) Therefore Monkey(x) and Cat(x) for the same x. Hence “Some monkeys are cats” is guaranteed true.4) C2 claims “No dog is a cat,” but S2 explicitly states the opposite inclusion (all dogs are cats). Therefore C2 is false.Verification / Alternative check:Draw three sets with Dogs entirely inside Cats and Monkeys overlapping Dogs. The overlap region automatically lies within Cats, confirming C1. C2 contradicts the diagram and S2.Why Other Options Are Wrong:“Only conclusion II” and “either I or II” are impossible because S2 affirms dogs are cats.“Neither” ignores the necessary overlap derived in steps 1–3.Common Pitfalls:Confusing universal inclusion with its converse, or overlooking that one witness from the “some” premise is enough to prove an existential conclusion.Final Answer:Only conclusion I follows.

Question 13

Syllogism — Determine the logically necessary conclusion(s): Statements: • All fruits are leaves. • Some fruits are grapes. Conclusions: I. Some leaves are grapes. II. All grapes are fruits.

Accepted Answer

Introduction / Context:This problem mixes a universal inclusion with a particular existence statement. The key is to transfer membership through a superset using “all,” while avoiding illicit conversion in the second conclusion.Given Data / Assumptions:S1: Fruits ⊆ Leaves.S2: Some Fruits are Grapes (Fruits ∩ Grapes ≠ ∅).Conclusions: C1 “Some Leaves are Grapes.” C2 “All Grapes are Fruits.”Concept / Approach:When X ⊆ Y and Some X are Z, those “some X” are simultaneously in Y, yielding Some Y are Z. However, “Some Fruits are Grapes” does not imply “All Grapes are Fruits”; that would be an illicit converse.Step-by-Step Solution:1) From S2, pick an element x with Fruit(x) and Grape(x).2) From S1, Fruit(x) ⇒ Leaf(x).3) Therefore Leaf(x) and Grape(x), proving C1: Some Leaves are Grapes.4) C2 claims Grapes ⊆ Fruits, which is not given. Grapes might include non-fruits in some models; hence C2 does not follow.Verification / Alternative check:Venn sketch: Fruits entirely inside Leaves; Grapes overlapping Fruits partly. The overlap witnesses C1. But Grapes may also extend outside Fruits, so C2 is not necessary.Why Other Options Are Wrong:Options including C2 assume a converse of S2; “Neither” ignores the guaranteed overlap for C1.Common Pitfalls:Assuming that because some Fruits are Grapes, therefore all Grapes are Fruits. “Some” never licenses “all.”Final Answer:Only Conclusion I follows.

Question 14

Syllogism — Assess the conclusions: Statements: • No man is a monkey. • Hari is a man. Conclusions: I. Hari is not a monkey. II. All men are not Hari.

Accepted Answer

Introduction / Context:This is a direct test of applying a universal negative to a named individual. Conclusion II is phrased ambiguously in everyday English and must be handled with care in formal reasoning.Given Data / Assumptions:S1: Man ∩ Monkey = ∅ (no man is a monkey).S2: Hari ∈ Man.Conclusions: C1 “Hari is not a monkey.” C2 “All men are not Hari.”Concept / Approach:From a universal negative, any individual in the first set cannot be in the second set. For C2, the phrase “All men are not Hari” usually reads as “No man is Hari,” which contradicts S2 (Hari is a man). Some exam setters intend it to mean “Not all men are Hari,” which is a trivial truth but not a syllogistic consequence stated by the premises. Standard keys therefore reject C2.Step-by-Step Solution:1) From S2, Hari is a Man.2) From S1, no Man is a Monkey.3) Therefore Hari cannot be a Monkey. C1 follows.4) C2, read as “No man is Hari,” contradicts S2; read as “Not all men are Hari,” it is a tautological remark not licensed by the premises as a syllogistic conclusion. Hence C2 does not follow.Verification / Alternative check:Instantiate any model where Men and Monkeys are disjoint and Hari is in Men. C1 holds in every such model; C2 fails on the strict interpretation and is not a standard syllogistic inference on the weak interpretation.Why Other Options Are Wrong:Including C2 elevates an ambiguous, non-syllogistic reading to necessity, which is incorrect.Common Pitfalls:Misreading natural-language quantifiers like “all … not” and ignoring the disjointness provided by universal negatives.Final Answer:Only Conclusion I follows.

Question 15

Syllogism — Choose which conclusions must be true: Statements: • All books are trees. • All trees are lions. Conclusions: I. All books are lions. II. All lions are books. III. All trees are books. IV. Some lions are books.

Accepted Answer

Introduction / Context:This is a textbook “chain” problem with two universal affirmatives. You must recognize transitivity of inclusion and then consider whether an existential claim (“some …”) is warranted.Given Data / Assumptions:S1: Books ⊆ Trees.S2: Trees ⊆ Lions.Conclusions: I “Books ⊆ Lions”; II “Lions ⊆ Books”; III “Trees ⊆ Books”; IV “Some Lions are Books.”Concept / Approach:From S1 and S2, Books ⊆ Lions by transitivity. The converses II and III are not implied. For IV, most exam conventions assume non-emptiness of the named classes; if any book exists, that book is a lion, yielding “Some lions are books.”Step-by-Step Solution:1) S1 + S2 ⇒ Books ⊆ Lions. So I follows.2) II (“All lions are books”) would need Lions ⊆ Books, not provided.3) III (“All trees are books”) would need Trees ⊆ Books, also not provided.4) If Books is non-empty (standard assumption), then ∃x Book(x). From S1 and S2, Book(x) ⇒ Lion(x), proving IV.Verification / Alternative check:Draw Books inside Trees, Trees inside Lions. The diagram makes I and IV immediate and shows II and III need not hold.Why Other Options Are Wrong:Any choice including II or III asserts a converse not supported by the premises; “none” ignores I (and often IV).Common Pitfalls:Assuming equivalence from inclusion and forgetting to verify existence for “some” conclusions.Final Answer:Only Conclusions I and IV follow.

Question 16

Syllogism — Judge the conclusions: Statements: • Some managers are young. • All boys are young. Conclusions: I. Some boys are managers. II. Some managers are boys.

Accepted Answer

Introduction / Context:Here two statements constrain who is “young,” but say nothing directly about overlap between Boys and Managers. The options tempt you to infer intersections that are not forced by the premises.Given Data / Assumptions:S1: Some Managers are Young.S2: All Boys are Young (Boys ⊆ Young).Conclusions: I “Some Boys are Managers.” II “Some Managers are Boys.”Concept / Approach:Two distinct subsets of a larger set (here, Young) need not overlap each other. The premises allow Managers and Boys to be disjoint subsets of Young, or to overlap; nothing forces an intersection.Step-by-Step Solution:1) Construct a model: Let Young be a big set. Place some Managers inside Young. Place all Boys inside Young, but in a region disjoint from Managers.2) Both premises hold (Managers ⊆ Young partly; Boys ⊆ Young entirely), yet neither I nor II is true in this model.3) Since the conclusions fail in a valid model, they are not necessary consequences.Verification / Alternative check:If you instead draw an overlap between Boys and Managers, I and II could become true—but “could be true” is not enough. We need “must be true.” Since countermodels exist, neither conclusion follows.Why Other Options Are Wrong:Any option claiming one or both follow confuses possibility with necessity.Common Pitfalls:Assuming that two sets included in a third must overlap, which is a common but invalid intuition.Final Answer:Neither Conclusion I nor Conclusion II follows.

Question 17

Syllogism — Identify which conclusions are compelled: Statements: • All astronomers are scientists. • Some scientists are shopkeepers. Conclusions: I. All astronomers are shopkeepers. II. Some shopkeepers are astronomers. III. Some shopkeepers are s…

Accepted Answer

Introduction / Context:We have one universal inclusion and one existential statement about scientists. We must filter out converses and overstatements, keeping only what the premises force.Given Data / Assumptions:S1: Astronomers ⊆ Scientists.S2: Some Scientists are Shopkeepers.Conclusions: I “Astronomers ⊆ Shopkeepers”; II “Some Shopkeepers are Astronomers”; III “Some Shopkeepers are Scientists”; IV “Scientists ⊆ Astronomers.”Concept / Approach:From S2 we immediately know there exist individuals who are both Scientists and Shopkeepers; that alone proves III. However, I and II require information linking Astronomers specifically to Shopkeepers, which we do not have. IV is a converse of S1 and is not supported.Step-by-Step Solution:1) Take x with Scientist(x) and Shopkeeper(x) from S2 ⇒ CIII holds.2) For I to hold, we would need Astronomers ⊆ Shopkeepers; the premises give no such link.3) For II to hold, some Shopkeeper would have to be an Astronomer; possible but not necessary.4) IV claims Scientists ⊆ Astronomers, the converse of S1; not implied.Verification / Alternative check:Build a model where Astronomers is a tiny subset of Scientists that does not overlap with Shopkeepers, while some other Scientists are Shopkeepers. Premises true, only III follows.Why Other Options Are Wrong:Options that include I, II or IV assert more than the premises provide or rely on illicit conversion.Common Pitfalls:Assuming that because some Scientists are Shopkeepers, subsets of Scientists (Astronomers) are also Shopkeepers; that is invalid without additional information.Final Answer:Only Conclusion III follows.

Question 18

Syllogism — Decide which conclusion(s) must hold: Statements: • Some principals are teachers. • All teachers are students. Conclusions: I. All principals are students. II. Some students are principals.

Accepted Answer

Introduction / Context:This problem combines a particular statement with a universal inclusion. The goal is to push the “some” individuals through the chain while avoiding unwarranted universal claims about all principals.Given Data / Assumptions:S1: Some Principals are Teachers.S2: All Teachers are Students (Teachers ⊆ Students).Conclusions: I “All Principals are Students.” II “Some Students are Principals.”Concept / Approach:The subset relation allows transfer: those principals who are teachers are certainly students. That yields an existential overlap between Students and Principals. However, nothing says all Principals are Teachers (or Students), so the universal I does not follow.Step-by-Step Solution:1) From S1 pick x with Principal(x) and Teacher(x).2) From S2, Teacher(x) ⇒ Student(x).3) Hence Student(x) and Principal(x), proving II (Some Students are Principals).4) I would require every Principal to be a Student, which is not guaranteed.Verification / Alternative check:Diagram: Teachers inside Students; Principals overlaps Teachers partially. The overlap witnesses II; there may be Principals outside Teachers/Students, so I is not necessary.Why Other Options Are Wrong:Any option including I assumes a universal inclusion not present in the premises.Common Pitfalls:Generalizing from “some” to “all.”Final Answer:Only Conclusion II follows.

Question 19

Syllogism — Which conclusion(s) necessarily follow? Statements: • All parrots are chicks. • All birds are chicks. Conclusions: I. Some birds are parrots. II. Some chicks are parrots.

Accepted Answer

Introduction / Context:Two universal statements place both Parrots and Birds wholly inside Chicks. The trap is to infer an overlap between Birds and Parrots, which is not enforced by the premises.Given Data / Assumptions:S1: Parrots ⊆ Chicks.S2: Birds ⊆ Chicks.Conclusions: I “Some Birds are Parrots.” II “Some Chicks are Parrots.”Concept / Approach:Two different subsets of the same superset need not intersect. Therefore C1 is not necessary. However, provided Parrots exist (standard assumption for such categories), then certainly some Chicks are Parrots, because all Parrots are Chicks.Step-by-Step Solution:1) From S1, if ∃x Parrot(x), then x is a Chick. This witnesses C2.2) For C1, we would need Parrots ∩ Birds ≠ ∅, which the premises do not guarantee. They could be disjoint subsets within Chicks.Verification / Alternative check:Diagram with Parrots and Birds as two non-overlapping circles inside Chicks: S1 and S2 both hold, C2 holds (pick any Parrot), while C1 fails.Why Other Options Are Wrong:Including C1 assumes an overlap not given; “none” ignores the existential that follows from S1 given non-emptiness of Parrots.Common Pitfalls:Believing subsets of the same superset must overlap (they need not).Final Answer:Only Conclusion II follows.

Question 20

Syllogism — From the premises below, select the conclusion(s) that must hold: Statements: • Some bags are pockets. • No pocket is a pouch. Conclusions: I. Some bags are not pouches. II. Some pockets are bags.

Accepted Answer

Introduction / Context:This item pairs an existential overlap with a universal negative about one of the sets. The question is designed so that both directions of the “some” intersection can be affirmed and a negative conclusion is forced for at least some bags.Given Data / Assumptions:S1: Some Bags are Pockets (Bags ∩ Pockets ≠ ∅).S2: No Pocket is a Pouch (Pockets ∩ Pouches = ∅).Conclusions: I “Some Bags are not Pouches.” II “Some Pockets are Bags.”Concept / Approach:From S1, take an element that is both Bag and Pocket. S2 then guarantees that element is not a Pouch. That single witness proves I. S1 also directly implies II because “Some Bags are Pockets” is symmetric as an intersection statement: there exist Pockets that are Bags.Step-by-Step Solution:1) Pick x with Bag(x) and Pocket(x) (by S1).2) From S2, Pocket(x) ⇒ not Pouch(x).3) Therefore Bag(x) and not Pouch(x) ⇒ “Some Bags are not Pouches” (I) holds.4) Because x is also a Pocket, “Some Pockets are Bags” (II) is witnessed by the same element.Verification / Alternative check:Venn diagram: Bags and Pockets overlap; Pockets and Pouches are disjoint. The overlap region is necessarily outside Pouches, establishing I, and symmetry of overlap confirms II.Why Other Options Are Wrong:Choices that keep only one conclusion ignore the dual implications from the same witness; “neither” contradicts the guaranteed overlap.Common Pitfalls:Missing that “some A are B” automatically implies “some B are A,” and overlooking how a universal negative eliminates membership in a third set.Final Answer:Both Conclusion I and Conclusion II follow.

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