With awk, as with perl, you'll have to wrap terms in //, but it can be done:

awk '(/term1/ && /term2/) || (/term1/ && xor(/term3/, /term4/))' 

Note that xor() which is a non-standard GNU extension is a bitwise xor, not a logical one. It's fine here as those /regex/ only ever return 0 or 1.

With other awk's you can define a logical xor() as a function:

function xor(a, b) {
  return (a || b && ! (a && b))
}

Or

function xor(a, b) {
  return ((a && !b) || (b && !a))
}

With awk, as with perl, you'll have to wrap terms in //, but it can be done:

awk '(/term1/ && /term2/) || (/term1/ && xor(/term3/, /term4/))' 

and we could factor out term1 to have:

awk '/term1/ && (/term2/ || xor(/term3/, /term4/))'

Note that xor() which is a non-standard GNU extension is a bitwise xor, not a logical one. It's fine here as those /regex/ only ever return 0 or 1.

With other awk's you can define a logical xor() as a function:

function xor(a, b) {
  return (a || b && ! (a && b))
}

Or

function xor(a, b) {
  return ((a && !b) || (b && !a))
}

Or

function xor(a, b) {
  return (a ? !b : b)
}

or we could just write the condition without a function:

awk '/term1/ && (/term2/ || (/term3/ ? !/term4/ : /term4/))'

With awk, as with perl, you'll have to wrap terms in //, but it can be done:

awk '(/term1/ && /term2/) || (/term1/ && xor(/term3/, /term4/))' 

With awk, as with perl, you'll have to wrap terms in //, but it can be done:

awk '(/term1/ && /term2/) || (/term1/ && xor(/term3/, /term4/))' 

Note that xor() which is a non-standard GNU extension is a bitwise xor, not a logical one. It's fine here as those /regex/ only ever return 0 or 1.

With other awk's you can define a logical xor() as a function:

function xor(a, b) {
  return (a || b && ! (a && b))
}

Or

function xor(a, b) {
  return ((a && !b) || (b && !a))
}