Suppose p ==> q is true and q ==> r is true. It is desirable to show that p ==> r cannot be false. Suppose that it is false. Then p is true and r is false (the only way that the implication can be false). Since q ==> r is true, by definition of implication, it cannot happen that q is true and r is false. The only way is that q must be false. But since p ==> q is true, again by definition of implication, it cannot happen that p is true and q is false. This means that p must be false. But p was true by the assumption about p ==> r being false. Therefore p is both true and false, a logical impossibility.
Therefore, assuming p ==> r is false, leads to an impossibility. This means that the implication p ==> r must be true.
The idea here is to evaluate pieces of the two expressions p and (q or r), (p and q) or r by its participating pieces. p, (q or r) are pieces of the first expression, (p and q), r are pieces of the second expression. The pieces in parentheses are evaluated in columns 4 and 6. The expressions (2+3) and (1+2) indicate what columns contain truth values that are used to determine the truth values in 4 and 6. Look at these as rows, not columns.
1
2
3
4
5
6
7
(1+2)
(2) (1)
(4+5)
(3+6)
p q
p ==>q
~q ~p
~q ==> ~p
3,6 equivalent (have same truth value)
T
T
T
F
F
T
T
T
F
F
T
F
F
T
F
T
T
F
T
T
T
F
F
T
T
T
T
T