Inferring from (∃x)Fx to (∃x)(∃x)Fx using existential generalization?Is formal logic unsuitable for...
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Inferring from (∃x)Fx to (∃x)(∃x)Fx using existential generalization?
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Inferring from (∃x)Fx to (∃x)(∃x)Fx using existential generalization?
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I was introduced to EG as follows (for some name 'a'): One can infer from Fa to (∃x)Fx. But today within a proof my professor posted he used EG to infer from (∃x)Fx to (∃x)(∃x)Fx. I'm not sure how that is supposed to work. It seems he's taking the 'Fx' within (∃x)Fx and using EG to change that to (∃x)Fx in which case we get (∃x)(∃x)Fx. But isn't the EG rule supposed to be going from Fa to (∃x)Fx? How does one go from Fx to (∃x)Fx?
logic
asked 5 hours ago
Emily RenoldsEmily Renolds
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I was introduced to EG as follows (for some name 'a'): One can infer from Fa to (∃x)Fx. But today within a proof my professor posted he used EG to infer from (∃x)Fx to (∃x)(∃x)Fx. I'm not sure how that is supposed to work. It seems he's taking the 'Fx' within (∃x)Fx and using EG to change that to (∃x)Fx in which case we get (∃x)(∃x)Fx. But isn't the EG rule supposed to be going from Fa to (∃x)Fx? How does one go from Fx to (∃x)Fx?
logic
asked 5 hours ago
Emily RenoldsEmily Renolds
111
New contributor
Emily Renolds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I was introduced to EG as follows (for some name 'a'): One can infer from Fa to (∃x)Fx. But today within a proof my professor posted he used EG to infer from (∃x)Fx to (∃x)(∃x)Fx. I'm not sure how that is supposed to work. It seems he's taking the 'Fx' within (∃x)Fx and using EG to change that to (∃x)Fx in which case we get (∃x)(∃x)Fx. But isn't the EG rule supposed to be going from Fa to (∃x)Fx? How does one go from Fx to (∃x)Fx?
logic
asked 5 hours ago
Emily RenoldsEmily Renolds
111
New contributor
Emily Renolds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I was introduced to EG as follows (for some name 'a'): One can infer from Fa to (∃x)Fx. But today within a proof my professor posted he used EG to infer from (∃x)Fx to (∃x)(∃x)Fx. I'm not sure how that is supposed to work. It seems he's taking the 'Fx' within (∃x)Fx and using EG to change that to (∃x)Fx in which case we get (∃x)(∃x)Fx. But isn't the EG rule supposed to be going from Fa to (∃x)Fx? How does one go from Fx to (∃x)Fx?
logic
logic
asked 5 hours ago
Emily RenoldsEmily Renolds
111
New contributor
Emily Renolds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Emily RenoldsEmily Renolds
111
New contributor
Emily Renolds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Emily RenoldsEmily Renolds
111
New contributor
Emily Renolds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Emily RenoldsEmily Renolds
111
asked 5 hours ago
Emily RenoldsEmily Renolds
111
111
New contributor
Emily Renolds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Emily Renolds is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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This is a lot simpler than it might seem. The details depend on how exactly the system that you're studying is set up but as an example let's look at the one defined in the Stanford Encyclopedia entry on classical logic.
In that entry, existential generalization (labeled ∃I) is defined as follows:
(∃I) For any closed term t, if Γ⊢θ(v|t) then Γ⊢∃vθ.
where θ(v|t) is defined as the result of substituting t for each free occurrence of v in θ. So suppose θ is Fx, then θ(x|a) would be Fa, and (∃I) tells us that if we've derived θ(x|a) (that is, Fa) we can also derive ∃xθ (that is, ∃xFx).
Here is the crucial part for your question: what happens when x does not occur free in θ? Then x cannot be replaced. That is, if θ is ∃xFx (in which x is bound) then θ(x|a) is also ∃xFx, since x does not occur free in θ. Remember that according to (∃I) we can derive ∃xθ if we have θ(x|a). So if θ is ∃xFx we can derive ∃xθ which is ∃x∃xFx.
In short, (∃I) lets you insert ∃x at the start of any derived formula in which x does not occur free.
answered 3 hours ago
EliranEliran
4,49821433
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This is a lot simpler than it might seem. The details depend on how exactly the system that you're studying is set up but as an example let's look at the one defined in the Stanford Encyclopedia entry on classical logic.
In that entry, existential generalization (labeled ∃I) is defined as follows:
(∃I) For any closed term t, if Γ⊢θ(v|t) then Γ⊢∃vθ.
where θ(v|t) is defined as the result of substituting t for each free occurrence of v in θ. So suppose θ is Fx, then θ(x|a) would be Fa, and (∃I) tells us that if we've derived θ(x|a) (that is, Fa) we can also derive ∃xθ (that is, ∃xFx).
Here is the crucial part for your question: what happens when x does not occur free in θ? Then x cannot be replaced. That is, if θ is ∃xFx (in which x is bound) then θ(x|a) is also ∃xFx, since x does not occur free in θ. Remember that according to (∃I) we can derive ∃xθ if we have θ(x|a). So if θ is ∃xFx we can derive ∃xθ which is ∃x∃xFx.
In short, (∃I) lets you insert ∃x at the start of any derived formula in which x does not occur free.
answered 3 hours ago
EliranEliran
4,49821433
add a comment |
This is a lot simpler than it might seem. The details depend on how exactly the system that you're studying is set up but as an example let's look at the one defined in the Stanford Encyclopedia entry on classical logic.
In that entry, existential generalization (labeled ∃I) is defined as follows:
(∃I) For any closed term t, if Γ⊢θ(v|t) then Γ⊢∃vθ.
where θ(v|t) is defined as the result of substituting t for each free occurrence of v in θ. So suppose θ is Fx, then θ(x|a) would be Fa, and (∃I) tells us that if we've derived θ(x|a) (that is, Fa) we can also derive ∃xθ (that is, ∃xFx).
Here is the crucial part for your question: what happens when x does not occur free in θ? Then x cannot be replaced. That is, if θ is ∃xFx (in which x is bound) then θ(x|a) is also ∃xFx, since x does not occur free in θ. Remember that according to (∃I) we can derive ∃xθ if we have θ(x|a). So if θ is ∃xFx we can derive ∃xθ which is ∃x∃xFx.
In short, (∃I) lets you insert ∃x at the start of any derived formula in which x does not occur free.
answered 3 hours ago
EliranEliran
4,49821433
add a comment |
This is a lot simpler than it might seem. The details depend on how exactly the system that you're studying is set up but as an example let's look at the one defined in the Stanford Encyclopedia entry on classical logic.
In that entry, existential generalization (labeled ∃I) is defined as follows:
(∃I) For any closed term t, if Γ⊢θ(v|t) then Γ⊢∃vθ.
where θ(v|t) is defined as the result of substituting t for each free occurrence of v in θ. So suppose θ is Fx, then θ(x|a) would be Fa, and (∃I) tells us that if we've derived θ(x|a) (that is, Fa) we can also derive ∃xθ (that is, ∃xFx).
Here is the crucial part for your question: what happens when x does not occur free in θ? Then x cannot be replaced. That is, if θ is ∃xFx (in which x is bound) then θ(x|a) is also ∃xFx, since x does not occur free in θ. Remember that according to (∃I) we can derive ∃xθ if we have θ(x|a). So if θ is ∃xFx we can derive ∃xθ which is ∃x∃xFx.
In short, (∃I) lets you insert ∃x at the start of any derived formula in which x does not occur free.
answered 3 hours ago
EliranEliran
4,49821433
This is a lot simpler than it might seem. The details depend on how exactly the system that you're studying is set up but as an example let's look at the one defined in the Stanford Encyclopedia entry on classical logic.
In that entry, existential generalization (labeled ∃I) is defined as follows:
(∃I) For any closed term t, if Γ⊢θ(v|t) then Γ⊢∃vθ.
where θ(v|t) is defined as the result of substituting t for each free occurrence of v in θ. So suppose θ is Fx, then θ(x|a) would be Fa, and (∃I) tells us that if we've derived θ(x|a) (that is, Fa) we can also derive ∃xθ (that is, ∃xFx).
Here is the crucial part for your question: what happens when x does not occur free in θ? Then x cannot be replaced. That is, if θ is ∃xFx (in which x is bound) then θ(x|a) is also ∃xFx, since x does not occur free in θ. Remember that according to (∃I) we can derive ∃xθ if we have θ(x|a). So if θ is ∃xFx we can derive ∃xθ which is ∃x∃xFx.
In short, (∃I) lets you insert ∃x at the start of any derived formula in which x does not occur free.
answered 3 hours ago
EliranEliran
4,49821433
edited 3 hours ago
answered 3 hours ago
EliranEliran
4,49821433
answered 3 hours ago
EliranEliran
4,49821433
answered 3 hours ago
EliranEliran
4,49821433
4,49821433
add a comment |
add a comment |
Emily Renolds is a new contributor. Be nice, and check out our Code of Conduct.
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Emily Renolds is a new contributor. Be nice, and check out our Code of Conduct.
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