📚 Year 10 OCR Philosophy: Quick Reference Handbook | 哲学关键公式定理速查手册
Philosophy may not have equations like physics, but it is built on rigorous logical structures, argument forms, and conceptual 'formulae' that shape every debate. This quick reference handbook brings together the essential logical rules, ethical principles, and classical arguments you are expected to know in Year 10 OCR Philosophy, presented as clear, reusable templates. Use it to check the validity of reasoning, unpack seemingly complex ideas, and build your own philosophical arguments with confidence.
哲学或许不像物理那样充满方程式，但它建立在严密的逻辑结构、论证形式和概念“公式”之上，这些构成了每一次思想交锋的基石。这本速查手册汇集了你在 10 年级 OCR 哲学课程中需要掌握的基本逻辑规则、伦理原则和经典论证，并以清晰、可复用的模板形式呈现。用它来检验推理的有效性，拆解看似复杂的理念，并自信地构建你自己的哲学论证。
1. Modus Ponens (Affirming the Antecedent) | 肯定前件式
One of the most fundamental valid argument forms. If the premises are true, the conclusion must be true.
最基本的有效论证形式之一。如果前提为真，结论必然为真。
P → Q
P
∴ Q
If it is raining, the ground is wet. (P → Q)
如果下雨，地面就会湿。(P → Q)
It is raining. (P)
正在下雨。(P)
Therefore, the ground is wet. (Q)
因此，地面是湿的。(Q)
2. Modus Tollens (Denying the Consequent) | 否定后件式
Another essential valid form that works by eliminating the consequent. If the consequent is false, the antecedent must also be false.
另一种核心的有效形式，通过否定后件来运作。如果后件为假，前件必然也为假。
P → Q
¬Q
∴ ¬P
If the switch is on, the light is lit. (P → Q)
如果开关开着，灯就会亮。(P → Q)
The light is not lit. (¬Q)
灯没亮。(¬Q)
Therefore, the switch is not on. (¬P)
因此，开关没开着。(¬P)
3. Hypothetical Syllogism | 假言三段论
This rule chains two conditionals together, allowing you to derive a direct connection between the first and third statements.
这条规则将两个条件陈述串联起来，使你能够得出第一个和第三个陈述之间的直接联系。
P → Q
Q → R
∴ P → R
If you study, you will pass. (P → Q)
如果你学习，你就会通过。(P → Q)
If you pass, you will graduate. (Q → R)
如果你通过，你就会毕业。(Q → R)
Therefore, if you study, you will graduate. (P → R)
因此，如果你学习，你就会毕业。(P → R)
4. Disjunctive Syllogism | 选言三段论
When one of two options is eliminated, the other must be true. Works only if the disjunction is exhaustive.
当两个选项中的一个被排除时，另一个必然为真。仅当该选言是穷尽的时才有效。
P ∨ Q
¬P
∴ Q
Either the battery is dead or the starter is broken. (P ∨ Q)
要么电池没电了，要么启动器坏了。(P ∨ Q)
The battery is not dead. (¬P)
电池没死。(¬P)
Therefore, the starter is broken. (Q)
因此，启动器坏了。(Q)
5. Reductio ad Absurdum (Proof by Contradiction) | 归谬法
Assume the opposite of what you want to prove, show it leads to an absurd or impossible consequence, and therefore conclude the original claim must hold.
假设你想要证明的论点的反面，展示其会导致荒谬或不可能的后果，从而得出原主张必然成立的结论。
Assume ¬P
Show ¬P → (Q ∧ ¬Q)
∴ P
To prove "all triangles have three sides", assume a triangle with four sides exists. That contradicts the definition of a triangle, so the original statement is true.
要证明“所有三角形都有三条边”，假设存在一个四条边的三角形。这与三角形的定义矛盾，所以原命题为真。
6. JTB Formula for Knowledge | 知识的 JTB 公式
Plato defined knowledge as justified true belief. All three components must be present for a belief to count as knowledge.
柏拉图将知识定义为得到证成的真信念。三个要素必须齐备，一个信念才能算作知识。
Knowledge = Belief + Truth + Justification
| Component | Description |
| Belief | You must actually hold the proposition to be true. |
| Truth | The proposition must correspond to reality. |
| Justification | You must have adequate reasons or evidence. |
中文：JTB 公式：知识 = 信念 + 真理 + 证成。信念：你必须实际认为该命题为真。真理：该命题必须符合现实。证成：你必须拥有充分的理由或证据。
7. Gettier Counter‑Example Structure | 盖蒂尔反例结构
Gettier cases show that JTB is not sufficient. The formula to construct one is: a justified belief that is true only by luck.
盖蒂尔案例表明 JTB 并不充分。构造这种案例的公式是：一个得到证成的信念，但它只是凭运气为真。
JTB + Lucky Truth ≠ Knowledge
Smith has strong evidence that Jones will get the job (justified). Smith also believes "the man who will get the job has ten coins in his pocket". Unknown to Smith, he himself will get the job and also has ten coins in his pocket. The belief is true but only by chance – not knowledge.
史密斯有充分证据相信琼斯会得到这份工作（证成）。他还相信“得到这份工作的人口袋里有十枚硬币”。史密斯不知道的是，他自己会得到这份工作，并且自己口袋里也有十枚硬币。这个信念为真，但纯属偶然——不是知识。
8. Utilitarian Calculus | 功利主义计算
Bentham's act utilitarianism evaluates actions by the total happiness produced. The 'formula' weighs seven factors.
边沁的行动功利主义通过产生的总幸福来评价行为。这个“公式”权衡七个因素。
Net Utility = Σ (Intensity × Duration × Certainty × Extent) – Pain
| Intensity | How strong is the pleasure? |
| Duration | How long will it last? |
| Certainty | How likely is it to occur? |
| Propinquity (remoteness) | How soon will it arrive? |
| Fecundity | Will it lead to further pleasures? |
| Purity | Will it be followed by pains? |
| Extent | How many people are affected? |
中文：净效用 = Σ（强度 × 持续时间 × 确定性 × 广度）– 痛苦。七个因素：强度、持续时间、确定性、临近性、繁殖性、纯度、广度。
9. Kant's Categorical Imperative (First Formulation) | 康德的绝对命令（第一公式）
Act only according to that maxim whereby you can at the same time will that it should become a universal law.
只按照你同时能够愿意它成为一条普遍法则的准则去行动。
Universalise your Maxim → Contradiction? → Impermissible
Test: Can I rationally will that everyone makes false promises when it suits them? If universalised, the institution of promising would collapse, so the maxim fails the test.
测试：我能否理性地愿意每个人在对自己有利时都做出虚假承诺？如果普遍化，承诺制度将会崩溃，因此该准则不能通过测试。
10. The Design Argument (Paley's Watchmaker) | 设计论证（佩利的手表匠）
If you find a watch on a heath, its intricate parts working together for a purpose compel you to infer a watchmaker. By analogy, the universe's order implies a designer.
如果你在荒野中发现一块手表，其精密部件为了一个目的协同运作，这迫使你推断必有一位制表匠。以此类推，宇宙的秩序暗示着一位设计者。
Order → Purpose → Intelligent Designer
Premise 1: Complex, purpose‑driven artifacts have designers.
前提 1：复杂、有目的的人造物都有设计者。
Premise 2: The universe is complex and appears purpose‑driven.
前提 2：宇宙是复杂的，并且看似有目的驱动。
Conclusion: Therefore, the universe probably has a designer.
结论：因此，宇宙很可能有一个设计者。
11. The Cosmological Argument (Aquinas' Second Way) | 宇宙论论证（阿奎那的第二路）
Every effect has a cause. An infinite regress of causes is impossible, so there must be a First Cause (God).
每个结果都有原因。原因的无限回溯是不可能的，所以必有一位第一因（上帝）。
Cause → Effect chain cannot be infinite ∴ Unmoved Mover
We observe causation in the world. (P)
我们在世界中观察到因果关系。(P)
Nothing can cause itself. (Q)
没有事物能够自因。(Q)
If there were no first cause, there would be no intermediate causes and no ultimate effect. (R)
如果没有第一因，就不会有中间原因，也不会有最终结果。(R)
Therefore, there must exist a first uncaused cause. (Conclusion)
因此，必存在一个最初的不被推动的推动者。（结论）
12. The Logical Problem of Evil (Inconsistent Triad) | 恶的逻辑问题（不一致三元组）
If God is all‑powerful, all‑knowing, and all‑loving, evil should not exist. The presence of evil creates a logical contradiction with these attributes.
如果上帝是全能的、全知的和全善的，恶就不应存在。恶的存在与这些属性构成了逻辑矛盾。
Omnipotence ∧ Omnibenevolence → ¬Evil
Evil exists
∴ ¬(Omnipotence ∧ Omnibenevolence)
Mackie's triad: God is omnipotent; God is wholly good; yet evil exists. The theist must give up at least one of these to avoid contradiction.
麦基的三元组：上帝是全能的；上帝是全然良善的；但恶存在。有神论者必须放弃至少其中一个以避免矛盾。
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