1 files changed, 4 insertions(+), 4 deletions(-)

M content/blog/on-decidability-and-the-mu-puzzle.md

@@ 77,14 77,14 @@ have 3 as a prime factor. We can express this more succinctly as
 ∀x. x ≠ 0 (mod 3) --> 2x ≠ 0 (mod 3)
 ```
 
-Thus `2*n` is not divisible by 3, and we can never construct the string `MU`
+Thus `2*n` is not divisible by 3, and we can never construct the string `MU`,
 starting from `MI`.
 
 
 # Characterizing all generatable strings
 
-Not only have we shown that `MU` is not constructible, but also any other string
-with a value divisible by 3, starting from `MI`. The question remains: which
+Not only have we shown that `MU` is not constructible, starting from `MI`, but
+also any other string with a value divisible by 3. The question remains: which
 strings can we generate? Is it possible to generate all other strings, i.e. all
 strings `Mx` such that `value(Mx) != 0 (mod 3)`?
 


@@ 99,7 99,7 @@ The answer turns out to be yes, using the following algorithm.
 It is always possible to apply step 1: the infinite sequence of strings
 generated by repeatedly applying rule 1 to `MI` has values `1, 2, 4, 8, 16, 32,
 ...`, generating all powers of 2. Taking these values modulo 3 we get `1, 2, 1,
-2, 1, 2, 1, 2, ...`, i.e. `2^i (mod 3)` is `1` if `i` is even, and 2 otherwise. Since
+2, 1, 2, 1, 2, ...`, i.e. `2^i (mod 3)` is `1` if `i` is even, and `2` otherwise. Since
 `value(Mx) != 0` by assumption, there always exists a longer string `My` such that
 `value(My) = value(Mx) (mod 3)`.