Proof rules in discrete math. When you write that a element is arbitrary, for example, "Let z be an arbitrary animal", does that automatically implies that you are using generalization rule as a proof?

metal1fc

metal1fc

Answered question

2022-09-04

Proof rules in discrete math
When you write that a element is arbitrary, for example, "Let z be an arbitrary animal", does that automatically implies that you are using generalization rule as a proof?

Answer & Explanation

Manuel Leach

Manuel Leach

Beginner2022-09-05Added 13 answers

Step 1
In mathematical practice, in order to prove that a formula ϕ is true for every element x, a standard move is to write :
"let t be an element"
and then prove ϕ [ t / x ] without using any information about t. Then ϕ [ t / x ] must be true regardless of what element t is, and so x ϕ is proved.
Thus, when starting an argument with "let t", we introduce a new term (constant or variable) considering it "generic"; if we prove ϕ, then we can apply the -introduction rule (also called : Generalization) and conclude with x ϕ.
The -introduction rule is :
if Γ ϕ, then Γ x ϕ, provided that x is not free in any formula of Γ.
Step 2
We have to take care about the "let t arbitrary". Consider this (wrong) derivation :
1) x = 0 --- assumed [x is not "arbitrary" !]
2) x ( x = 0 ) --- Generalization (wrong : x is not free in 1)
3) x = 0 x ( x = 0 )
4) x [ x = 0 x ( x = 0 ) ] --- Generalization (now is "legal")
5) 0 = 0 x ( x = 0 ) --- from 4) by Instantiation.
The last formula is clearly false in N; thus, the derivation must be wrong !
An argument starting with "let t arbitrary" can also invlove -elimination:
if Γ , ϕ [ t / x ] ψ , then Γ , x ϕ ( x ) ψ , provided that t is not free in ψ or any formula of Γ.
In this case, we "know" that x ϕ and we we start by writing :
"let t be a ϕ s";
if from this assumption we can derive a conclusion ψ that does not mention t, we are licensed to conclude ψ from the premise : "there exists some ϕ s".

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