learn-prolog-now-exercises

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Hi, I think the correct answer is 7 clauses and 5 (not 7!) predicates.

upgrd

In chapter 6 it is possible to create a swapfl/2 using only the append/3 rule.

phochste

``` cd chapter-11 swipl practical-session.pl powerset([a,b,c,d,e],P). P = [[], [a], [a, b], [a, b, c], [a, b, c, d], [a, b, c|...], [a, b|...], [a|...], [...|...]|...]. ```

lengyijun

Chapter 3 exercise Exercise 3.3

1

Are you sure about the solution? It looks like OUT OF LOCAL STACK will be occur.

kagankorkmaz

RobbeB32

Hi, The solution provided will asset that subset([a,b,c,a], [a, b, c]). is true. It is missing checking that the first parameter is a set itself. Here I propose a fix...

nicoabie

I belive there is a bug. The current definition has the following outputs: ``` ?- display(p or r and s). or(p,and(r,s)) true. ``` Which is ok but: ``` ?- display(p...

nicoabie

Ex 4.2 #7 is probably correct definition

2

I tried the following `assert([[1,2]|4]).` and it returned true. Why is it labelled as syntactically incorrect?

BrazilForever11

Specifically, it uses the `not/1` predicate, which is not given in the text, and negation in general is not discussed until chapter 10. Unfortunately, I don't think there's a way...

TV4Fun

Changed the solution as not/1 predicate has not been introduced yet and I think should not be used.

SmitSanghvi94