Bacterius said:Hint: a set of properties {P1, P2, ...} is said to imply a conclusion C if C holds whenever all of {P1, P2, ...} hold. So to show that these properties do not in fact imply C amounts to finding a counterexample, that is, a situation where C holds but not all of P{1, P2, ...} hold. Specifically if you can find examples where {P1, P2, ...} all hold and C holds, and also examples where C holds but not all of {P1, P2, ...} hold, then that logically means that {P1, P2, ...} cannot on their own imply C - that is, the two are logically independent.
In this case {P1, P2, ...} are just symmetry and transition (transitivity?) and C is the property of reflexion (reflexivity?). Can you complete?
Equality is reflexive, so the fact that equality is reflexive is implied by anything: If $B$ is true, then $A\to B$ is also true regardless of $B$.solakis said:How can we prove that the two properties of equality, namely,symmetry and transition do not imply that of reflexion??
A counterexample to $P_1\land\dots\land P_n\to C$ would be a situation where all $P_i$ hold, but $C$ does not.Bacterius said:Hint: a set of properties {P1, P2, ...} is said to imply a conclusion C if C holds whenever all of {P1, P2, ...} hold. So to show that these properties do not in fact imply C amounts to finding a counterexample, that is, a situation where C holds but not all of P{1, P2, ...} hold.
The second set of examples would show that $C$ does not imply all of $P_i$, but it wouldn't show whether $P_1\land\dots\land P_n$ implies $C$.Bacterius said:Specifically if you can find examples where {P1, P2, ...} all hold and C holds, and also examples where C holds but not all of {P1, P2, ...} hold, then that logically means that {P1, P2, ...} cannot on their own imply C
Evgeny.Makarov said:Equality is reflexive, so the fact that equality is reflexive is implied by anything: If $B$ is true, then $A\to B$ is also true regardless of $B$.
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Evgeny.Makarov said:A counterexample to $P_1\land\dots\land P_n\to C$ would be a situation where all $P_i$ hold, but $C$ does not.
The second set of examples would show that $C$ does not imply all of $P_i$, but it wouldn't show whether $P_1\land\dots\land P_n$ implies $C$.
Briefly, yes. If you want to go deeper, we should start by clarifying terminology, in particular, the meaning of transition and reflexion (post #2 has proper terms for what you likely have in mind).solakis said:You mean that symmetry and transition imply reflexion in equality?
Then the following is provable:Evgeny.Makarov said:Briefly, yes. If you want to go deeper, we should start by clarifying terminology, in particular, the meaning of transition and reflexion (post #2 has proper terms for what you likely have in mind).
Yes.solakis said:Then the following is provable:
{\(\displaystyle (\forall A \forall B[A=B \Longrightarrow B=A])\wedge (\forall A \forall B \forall C [(A=B \wedge B=C)\Longrightarrow A=C])\)} \(\displaystyle \Longrightarrow \forall A[A=A]\)
Clarifying terms means making their meaning clear. The meaning of the terms "transition" and "reflexion", when they are used in the context of relations, is unclear to me. Please reread what others have said about these terms in this thread.solakis said:I don't understand what you mean by clarifying terms... etc etc
Evgeny.Makarov said:Yes.
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Evgeny.Makarov said:For this I need the logical calculus (Hilbert style, natural deduction, natural deduction in Fitch style, sequent calculus, etc.) and axioms.
I expect that there is an axiom $\forall x.\,x=x$. It is a part of all so-called theories with equality. Then your formula is obtained by simply adding vacuous assumptions.
Evgeny.Makarov said:I don't agree, and you have not provided any reasons for your opinion.
I agree that it is an axiom, and that's what I said back in post #4 and repeated in post #11. If you are talking about equality, then it comes with axioms ensuring at least that it is an equivalence relation and in particular reflexive. If you are talking about some arbitrary relation, then, indeed, symmetry and transitivity do not imply reflexivity. A counterexample is the relation $\{(2,2),(3,3),(2,3),(3,2)\}$ on the set $\{1,2,3\}$. However, if a relation $R$ on $X$ is left-total in addition to being symmetric and transitive, then it is reflexive. The fact that $R$ is left-total means $\forall x\in X\,\exists y\in Y\;xRy$. In the counterexample above, 1 is not related to any number, so the relation is not left-total.solakis said:Hence a=a for all a , is not a theorem implied by the other two properties of equality ,thus it may be considered as an axiom
The relationship betweensolakis said:Well, in the previous formula if we put A=B=C=a we have the following formula:
{(a=a=> a=a)&[ (a=a)&(a=a)=>a]} =>a=a
You probably mean a tautology, not an identity.solakis said:Is this formula an indentity ? NO
Yes you are right, i agree.solakis said:Well, in the previous formula if we put A=B=C=a we have the following formula:
{(a=a=> a=a)&[ (a=a)&(a=a)=>a]} =>a=a
Is this formula an indentity ? NO
Because if you take a=a to be false then you have : T=>F which is false.
Hence a=a for all a , is not a theorem implied by the other two properties of equality ,thus it may be considered as an axiom
Symmetry in proving inequality refers to the idea that if one side of an inequality is changed, the other side must also be changed in the same way. This maintains equality and ensures that the inequality remains valid.
Transition involves manipulating both sides of an inequality in the same way to maintain equality. This could involve adding or subtracting the same value, multiplying or dividing by the same number, or applying the same function to both sides. The key is to ensure that the inequality remains valid throughout the transition.
Reflexion, also known as reflection, is a type of transformation that involves flipping a figure or equation over a line of symmetry. In terms of proving inequality, this means that both sides of the inequality must be mirrored over the line of symmetry to maintain equality and validity.
Sure! Let's say we have the inequality 3x+5 < 2x+10. We can use symmetry by subtracting 2x from both sides to get x+5 < 10. Then, we can use transition by subtracting 5 from both sides to get x < 5. Finally, we can use reflexion by reflecting the inequality over the line x=5 to get 5 > x, which is equivalent to x < 5. Thus, we have proved the inequality using symmetry, transition, and reflexion.
Yes, there are various other methods for proving inequality such as using graphs, substitution, or induction. Each method may be more suitable for different types of inequalities or situations, so it's important to be familiar with multiple techniques for proving inequality.