设q(x)是关于类型为T的x的对象的可证明属性，那么q(y)对于类型为S的对象y应该是可证明的，其中S是T的子类型。

实际上，公认的答案并不是利斯科夫原理的反例。正方形自然是一个特定的矩形，因此从类矩形继承是完全有意义的。你只需要以这样的方式实现它:

@Override
public void setHeight(double height) {
   this.height = height;
   this.width = height; // since it's a square
}

@Override
public void setWidth(double width) {
   setHeight(width);
}

所以，提供了一个很好的例子，然而，这是一个反例:

class Family:
-- getChildrenCount()

class FamilyWithKids extends Family:
-- getChildrenCount() { return childrenCount; } // always > 0

class DeadFamilyWithKids extends FamilyWithKids:
-- getChildrenCount() { return 0; }
-- getChildrenCountWhenAlive() { return childrenCountWhenAlive; }

在这个实现中，DeadFamilyWithKids不能从FamilyWithKids继承，因为getChildrenCount()返回0，而从FamilyWithKids它应该总是返回大于0的值。

罗伯特·马丁有一篇关于利斯科夫替换原理的优秀论文。它讨论了可能违反原则的微妙和不那么微妙的方式。

论文的一些相关部分(注意，第二个例子被大量压缩):

A Simple Example of a Violation of LSP One of the most glaring violations of this principle is the use of C++ Run-Time Type Information (RTTI) to select a function based upon the type of an object. i.e.: void DrawShape(const Shape& s) { if (typeid(s) == typeid(Square)) DrawSquare(static_cast<Square&>(s)); else if (typeid(s) == typeid(Circle)) DrawCircle(static_cast<Circle&>(s)); } Clearly the DrawShape function is badly formed. It must know about every possible derivative of the Shape class, and it must be changed whenever new derivatives of Shape are created. Indeed, many view the structure of this function as anathema to Object Oriented Design. Square and Rectangle, a More Subtle Violation. However, there are other, far more subtle, ways of violating the LSP. Consider an application which uses the Rectangle class as described below: class Rectangle { public: void SetWidth(double w) {itsWidth=w;} void SetHeight(double h) {itsHeight=w;} double GetHeight() const {return itsHeight;} double GetWidth() const {return itsWidth;} private: double itsWidth; double itsHeight; }; [...] Imagine that one day the users demand the ability to manipulate squares in addition to rectangles. [...] Clearly, a square is a rectangle for all normal intents and purposes. Since the ISA relationship holds, it is logical to model the Square class as being derived from Rectangle. [...] Square will inherit the SetWidth and SetHeight functions. These functions are utterly inappropriate for a Square, since the width and height of a square are identical. This should be a significant clue that there is a problem with the design. However, there is a way to sidestep the problem. We could override SetWidth and SetHeight [...] But consider the following function: void f(Rectangle& r) { r.SetWidth(32); // calls Rectangle::SetWidth } If we pass a reference to a Square object into this function, the Square object will be corrupted because the height won’t be changed. This is a clear violation of LSP. The function does not work for derivatives of its arguments. [...]

LSP是关于类的契约的规则:如果基类满足契约，则LSP派生的类也必须满足该契约。

在Pseudo-python

class Base:
   def Foo(self, arg): 
       # *... do stuff*

class Derived(Base):
   def Foo(self, arg):
       # *... do stuff*

如果每次在派生对象上调用Foo，它给出的结果与在Base对象上调用Foo完全相同，只要arg是相同的。

在一个非常简单的句子中，我们可以说:

子类不能违背它的基类特征。它必须有能力。我们可以说这和子类型是一样的。

我建议您阅读这篇文章:违反利斯科夫替换原则(LSP)。

你可以在那里找到一个解释，什么是利斯科夫替换原则，一般线索帮助你猜测你是否已经违反了它，一个方法的例子，将帮助你使你的类层次结构更安全。

利科夫替换原则指出，如果程序模块使用基类，则基类的引用可以被派生类替换，而不会影响程序模块的功能。

派生类型必须能够完全替代它们的基类型。

示例- java中的协变返回类型。