7.真值表法

例2
如果爆發流感 (P)，則學校停課 (Q)。 如果學校停課，則學生很高興 (R)。 所以，如果爆發流感，則學生很高興。
P→Q Q→R P→R
Indirect truth tables
步驟：
要檢驗一個argument是否valid，先假設它 是invalid，即假設前提皆真而結論假。
然後以倒溯的方法求各式的真值，過程中 若出現矛盾，則此argument是valid；若不 出現矛盾，則此argument是invalid.
Testing statements for consistency
1. Begin by writing the statements on a line, separating each with “ / ”. 2. Then assume that the statements are consistent. (Assign a T to the main operator of each.) 3. Compute the truth values of the components. If it leads to a contradiction, then the statements are inconsistent. Otherwise, they are consistent.
Contingency
(P∧〜Q) → R A compound statement is said to T F F T T T TF FT T F be contingent if its TT TF T T truth value varies TT TF F F depending on the FF FT T T truth values of its FF FT T F components.
FF TF T T FF TF T F
Logically equivalent
A→B T T T T F F F T T F T F 〜B → 〜A FT T FT TF F FT FT T TF TF T TF
Contradictory
A→B T T T T F F F T T F T F A∧〜B TF FT TT TF FF FT FF TF
例1
〜P → (Q∨R) 〜Q R→P
將公式並列： 〜P → (Q∨R) / 〜Q // R → P
例2
A → (B∨C) B→D A 〜C → D A → (B∨C) / B → D / A // 〜C → D
例3
〜A → B B→A A→〜B A∧〜B
〜A → B / B → A / A→〜B // A∧〜B
Only if all three lines had led to a contradiction would these statements be inconsistent.
例1
P∨Q / Q → (R∨P) / R → 〜Q / 〜P FTT TT FFF F T FT TF Contradictory! So, the statements are inconsistent.
例2
P → (Q∧R) / R → 〜P / Q∨P / Q → R F T TTT TT TF TTF T T T F T T F T F No contradiction. So, the statements are consistent.
例1
If John studies hard, he will pass the exam. John does not study hard. Therefore, he will not pass the exam.
J: John studies hard. P: John will pass the exam.
策略
If an indirect truth table requires more than one line, the method to be followed is this. Either select one of the premises and compute all of the ways it can be made true, or select the conclusion and compute all of the ways it can be made false. This selection should be dictated by the requirement of simplicity.
Note
As with testing argument, the objective is to avoid a contradiction. As soon as no contradiction is reached, we stop. The statements are consistent.
真值表法
A truth table gives the truth value of a compound proposition for every possible truth value of its simple components.
真值表就是列出所有的真假的組合。 （當字母的種類是n時，真值表的行數L=2ⁿ）
一個字母
若只有一種字母，則只 有兩種可能： T， F。
A T F
兩個字母
兩種字母則有四種可能 的組合。 先在A下面寫下： T T F F。 再在B下面寫下： T F T F。 A T T F F B T F T F
三個字母
三種字母則有八種可能 的組合。
先在A下面寫下： T T T T F F F F。 然後在B下面寫下： T T F F T T F F。 最後在C下面寫下： T F T F T F T F。
Contradiction (矛盾)
A compound (A∨B)≡(〜A∧〜B) statement is said to T T T F F T F F T be logically false or T T F F F T F T F self-contradictory if F T T F T F F F T it is false FFF F TFT TF regardless of the truth values of its components.
Consistent
A∨B TTT TTF FTT FFF A∧B TTT TFF FFT FFF
Inconsistent
A≡B TTT TFF FFT FTF A∧〜B TFFT TTTF FFFT FFTF
Truth tables for argument
步驟：
1. 將論證用符號表達。 2. 將表達式並列，用 “ / ” 隔開前提，用 “ // ” 隔開結論。 3. 分別列出真值表。 4. 看看有沒有所有前提皆真而結論假的情形， 若有，則論證無效；若無，則有效。
A T T T T F F F F
B T T F F T T F F
C T F T F T F T F
例子
(P∨〜Q) → Q
Βιβλιοθήκη BaiduP∧〜Q) → R
Tautology（重言句）
例子 A compound statement is said to be logically true or ((A → B)∧A) → B tautologous if it is T T T T T T T true regardless of T F F FT T F the truth values of its components. F T T FF T T F T F FF T F