Modus Ponens on A Statement of Deduction?

  • Thread starter darkchild
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In summary, the theorem being proven in this conversation is that if a set of formulas, represented by Δ, can prove a universally quantified formula ##\forall vP##, then it can also prove the formula ##P(t/v)##, provided that the predicate ##P## can accept the term ##t## in place of the variable ##v##. The confusion lies in the application of modus ponens in the proof, where the symbol Δ may seem to be ignored, but is actually being used in a formal way to show that the deduction of ##\forall vP## from Δ can lead to a deduction of ##P(t/v)## from Δ.
  • #1
darkchild
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This is from a text on mathematical logic. The theorem to be proven (specialization):

If Δ [itex]\vdash[/itex] [itex]\forall[/itex]vP, then Δ [itex]\vdash[/itex] P(t/v), provided that P admits t for v.

My confusion concerns the use of modens ponens in the proof:

Suppose that Δ [itex]\vdash[/itex] [itex]\forall[/itex]vP and P admits t for v. Then modus ponens applied to Δ [itex]\vdash[/itex] [itex]\forall[/itex]vP and [itex]\vdash \forall[/itex]vP [itex]\rightarrow[/itex]P(t/v) (Axiom Scheme A5) gives Δ [itex]\vdash[/itex] P(t/v).

I have never seen this before and do not understand how it is legal or exactly what it means to use modus ponens on statements containing Δ (a set of formulas used as premises) and [itex]vdash[/itex]. It seems the latter are simply ignored, yet they are crucial to the meaning of the statement.
 
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  • #2
darkchild said:
This is from a text on mathematical logic. The theorem to be proven (specialization):

If Δ [itex]\vdash[/itex] [itex]\forall[/itex]vP, then Δ [itex]\vdash[/itex] P(t/v), provided that P admits t for v.

My confusion concerns the use of modens ponens in the proof:

Suppose that Δ [itex]\vdash[/itex] [itex]\forall[/itex]vP and P admits t for v. Then modus ponens applied to Δ [itex]\vdash[/itex] [itex]\forall[/itex]vP and [itex]\vdash \forall[/itex]vP [itex]\rightarrow[/itex]P(t/v) (Axiom Scheme A5) gives Δ [itex]\vdash[/itex] P(t/v).

I have never seen this before and do not understand how it is legal or exactly what it means to use modus ponens on statements containing Δ (a set of formulas used as premises) and [itex]vdash[/itex]. It seems the latter are simply ignored, yet they are crucial to the meaning of the statement.

I looks like what is intended is that the deduction of ##\forall Pv## from ##\Delta## and modus ponens applied to ##\forall Pv## and ##\forall Pv\rightarrow P(t/v)## gives a deduction of ##P(t/v)## from ##\Delta##. I would agree that the wording is a bit wonky, though.
 
  • #3
What I don't understand is the role played by the symbol Δ. I understand modus ponens as

1. S
2. S → T
∴ T

That cannot be neatly applied to this proof:

1. Δ [itex]\vdash \forall[/itex]vP would correspond to S
2. [itex]\vdash[/itex][itex]\forall[/itex]vPP(t/v) should correspond to S → T

However, the formula that corresponds to S in step one is different than the formula that corresponds to S in step 2. It's missing Δ.
 
  • #4
darkchild said:
What I don't understand is the role played by the symbol Δ. I understand modus ponens as

1. S
2. S → T
∴ T

That cannot be neatly applied to this proof:

1. Δ [itex]\vdash \forall[/itex]vP would correspond to S
2. [itex]\vdash[/itex][itex]\forall[/itex]vPP(t/v) should correspond to S → T

However, the formula that corresponds to S in step one is different than the formula that corresponds to S in step 2. It's missing Δ.

##\forall vP## corresponds to ##S## and ##P(t/v)## corresponds to ##T##.

Again, the wording of the text is a little bit off in my opinion. They aren't applying "informal" modus ponens to the meta-mathematical statements ##\Delta\vdash \forall vP## and ##\vdash \left(\forall vP\rightarrow P(t/v)\right)##; they're applying "formal" modus ponens to the formal sentences ##\forall vP## and ##\forall vP\rightarrow P(t/v)##.

Note that this is an informal proof that there is a formal deduction of ##P(t/v)## from ##\Delta## given the fact that there is a formal deduction of ##\forall vP## from ##\Delta##. It's basically a proof about proofs, and it's more than a little bit meta.
 
  • #5
Can you explain?Modus Ponens is a rule of inference that allows us to draw a conclusion from a conditional statement and its antecedent. In this case, the conditional statement is \forallvP \rightarrow P(t/v) and the antecedent is \forallvP.

To understand why this is a valid use of Modus Ponens, we need to look at the Axiom Scheme A5, which states that if P admits t for v, then \forallvP \rightarrow P(t/v) is a valid formula. This means that if P admits t for v, then the conditional statement \forallvP \rightarrow P(t/v) is true.

Now, since we have assumed that \forallvP is true (since we have \Delta \vdash \forallvP), we can use Modus Ponens to infer that P(t/v) is also true. This is because the antecedent (\forallvP) is true and the conditional statement (\forallvP \rightarrow P(t/v)) is also true, so we can conclude that the consequent (P(t/v)) must also be true.

In other words, Modus Ponens allows us to infer P(t/v) from the premises \Delta \vdash \forallvP and \vdash \forallvP \rightarrow P(t/v). The use of \Delta in the proof is simply to indicate that \forallvP is one of the premises in the set \Delta, and it is used to show that \forallvP is true.

Therefore, using Modus Ponens in this way is completely valid and necessary for the proof of the theorem. It allows us to draw a conclusion from the premises and the Axiom Scheme A5, which is crucial for proving the specialization theorem.
 

1. What is "Modus Ponens" in deductive reasoning?

Modus Ponens is a rule of inference in deductive reasoning that states if the antecedent (if-portion) of a conditional statement is true, then the consequent (then-portion) must also be true.

2. How does "Modus Ponens" work in a statement of deduction?

In a statement of deduction, "Modus Ponens" is used to conclude that the consequent of a conditional statement is true based on the truth of the antecedent and the statement itself.

3. What is the logical form of "Modus Ponens" in a statement of deduction?

The logical form of "Modus Ponens" is:

If P implies Q and P is true, then Q is true.

4. Can "Modus Ponens" be used in reverse?

No, "Modus Ponens" cannot be used in reverse. It only works in one direction, from the antecedent to the consequent.

5. What are some common examples of "Modus Ponens" in real life?

One common example of "Modus Ponens" in real life is: "If it rains, the ground will be wet. It is raining. Therefore, the ground is wet."

Another example is: "If I study for the exam, I will pass. I studied for the exam. Therefore, I will pass."

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