Why is this Statement correct: G implies ¬Contradiction?

Question

_{Source: p 110. Sweet Reason: A Field Guide to Modern Logic (2010 2 ed) by Henle, Garfield, Tymoczko.}

The following statements are all about schools. Here is some partial information:
A is true at Sophist College
B is false at Stoic University.
C is consistent (i.e. its true at some school).
D is contingent.
E is a contradiction.
F implies B.
[No information is supplied about G.]

For each of the following statements respond Y if the statement is correct, N if the statement is incorrect, or I if there is not enough information to decide whether or not it is correct.

17. G implies ¬E.

[Answer on p 357, and online for Section 5.3 too:] 17. Y

My initial answer was I: we do not know the truth value of G or ¬E, because per the following definition, ¬E means that not every line of E's truth table is false (= ≥ 1 line is true).

[p 109:] If a wff [Well-formed Formula] is false on every line of its truth table we say it's a contradiction.

So I was stupefied to see the answer as Y.

Answer 1

In the truth table of a contradiction E, all lines are false. Then, in the truth table of ¬E, all lines must be ¬false, i.e. true.

Since G → ¬E ≡ ¬G ∨ ¬E, and we know that ¬E is always true, the implication must be as well.

You were wrong in deriving from the quoted definition that

¬E means that not every line of E's truth table is false (= ≥ 1 line is true)

You could say that a statement is not a contradiction iff there is at least one line true. However, this is a meta-theorem about propositions and truth tables, which does not give you the truth table of a negated proposition.

Answer 2

A statement of the form P → Q where Q is true is always true, regardless of the truth value of P: it is said to be trivially true.