Inferring from (∃x)Fx to (∃x)(∃x)Fx using existential generalization?
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 Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
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 Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
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 Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
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 Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
asked Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
asked Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
asked Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
asked Mar 3 at 3:14
Emily RenoldsEmily Renolds
261
261
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 Mar 3 at 5:04
EliranEliran
4,89331433
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "265"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphilosophy.stackexchange.com%2fquestions%2f60866%2finferring-from-%25e2%2588%2583xfx-to-%25e2%2588%2583x%25e2%2588%2583xfx-using-existential-generalization%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Mar 3 at 5:04
EliranEliran
4,89331433
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 Mar 3 at 5:04
EliranEliran
4,89331433
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 Mar 3 at 5:04
EliranEliran
4,89331433
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 Mar 3 at 5:04
EliranEliran
4,89331433
edited Mar 3 at 5:28
answered Mar 3 at 5:04
EliranEliran
4,89331433
answered Mar 3 at 5:04
EliranEliran
4,89331433
answered Mar 3 at 5:04
EliranEliran
4,89331433
4,89331433
add a comment |
add a comment |
Thanks for contributing an answer to Philosophy Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphilosophy.stackexchange.com%2fquestions%2f60866%2finferring-from-%25e2%2588%2583xfx-to-%25e2%2588%2583x%25e2%2588%2583xfx-using-existential-generalization%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
