Why does “false && true” not exit with set -e in Bash? [duplicate]

Answer 1

true && false; echo success returns the return code from echo which is 0.

 bash -c 'false && true'; echo $?

yields 1 as expected. Pheew!

Aside: bash -c 'set -e; false || false; echo success'; echo $? yields return code 1 because second command does not execute echo (stops at false because errors stop the current command), note that success is not printed.