Removing specific rows from a dataframe

DF[ ! ( ( DF$sub ==1 & DF$day==2) | ( DF$sub ==3 & DF$day==4) ) , ]   # note the ! (negation)

Or if sub is a factor as suggested by your use of quotes:

DF[ ! paste(sub,day,sep="_") %in% c("1_2", "3_4"), ]

Could also use subset:

subset(DF,  ! paste(sub,day,sep="_") %in% c("1_2", "3_4") )

(And I endorse the use of which in Dirk's answer when using "[" even though some claim it is not needed.)

This boils down to two distinct steps:

1. Figure out when your condition is true, and hence compute a vector of booleans, or, as I prefer, their indices by wrapping it into which()
2. Create an updated data.frame by excluding the indices from the previous step.

Here is an example:

R> set.seed(42)
R> DF <- data.frame(sub=rep(1:4, each=4), day=sample(1:4, 16, replace=TRUE))
R> DF
   sub day
1    1   4
2    1   4
3    1   2
4    1   4
5    2   3
6    2   3
7    2   3
8    2   1
9    3   3
10   3   3
11   3   2
12   3   3
13   4   4
14   4   2
15   4   2
16   4   4
R> ind <- which(with( DF, sub==2 & day==3 ))
R> ind
[1] 5 6 7
R> DF <- DF[ -ind, ]
R> table(DF)
   day
sub 1 2 3 4
  1 0 1 0 3
  2 1 0 0 0
  3 0 1 3 0
  4 0 2 0 2
R> 

And we see that sub==2 has only one entry remaining with day==1.

Edit The compound condition can be done with an 'or' as follows:

ind <- which(with( DF, (sub==1 & day==2) | (sub=3 & day=4) ))

and here is a new full example

R> set.seed(1)
R> DF <- data.frame(sub=rep(1:4, each=5), day=sample(1:4, 20, replace=TRUE))
R> table(DF)
   day
sub 1 2 3 4
  1 1 2 1 1
  2 1 0 2 2
  3 2 1 1 1
  4 0 2 1 2
R> ind <- which(with( DF, (sub==1 & day==2) | (sub==3 & day==4) ))
R> ind
[1]  1  2 15
R> DF <- DF[-ind, ]
R> table(DF)
   day
sub 1 2 3 4
  1 1 0 1 1
  2 1 0 2 2
  3 2 1 1 0
  4 0 2 1 2
R>