Natural deduction without premises given?
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1
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Normally when given a question like $Q wedge P, R vdash P wedge R$
I can do box proof like:
$dfrac{dfrac{Q wedge P^{~text{(assumption)}}}{P}{^text{($wedge$-elimination)}}quad R^{~text{(assumption)}}}{Pwedge R}{^text{($wedge$ introduction)}}
\text{ (Q.E.D.)}$
But what about when I'm asked to prove $A rightarrow (B rightarrow A)$?
Do I just start with assumption as $A$?
logic natural-deduction
asked Nov 13 at 21:17
Ansar Al
414
add a comment |
up vote
1
down vote
favorite
Normally when given a question like $Q wedge P, R vdash P wedge R$
I can do box proof like:
$dfrac{dfrac{Q wedge P^{~text{(assumption)}}}{P}{^text{($wedge$-elimination)}}quad R^{~text{(assumption)}}}{Pwedge R}{^text{($wedge$ introduction)}}
\text{ (Q.E.D.)}$
But what about when I'm asked to prove $A rightarrow (B rightarrow A)$?
Do I just start with assumption as $A$?
logic natural-deduction
asked Nov 13 at 21:17
Ansar Al
414
3
You use the $to$ introduction rule, which means you start with a temporary assumption of $A$ and derive $Bto A$ (also using $to$-introduction).
– spaceisdarkgreen
Nov 13 at 21:39
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Normally when given a question like $Q wedge P, R vdash P wedge R$
I can do box proof like:
$dfrac{dfrac{Q wedge P^{~text{(assumption)}}}{P}{^text{($wedge$-elimination)}}quad R^{~text{(assumption)}}}{Pwedge R}{^text{($wedge$ introduction)}}
\text{ (Q.E.D.)}$
But what about when I'm asked to prove $A rightarrow (B rightarrow A)$?
Do I just start with assumption as $A$?
logic natural-deduction
asked Nov 13 at 21:17
Ansar Al
414
Normally when given a question like $Q wedge P, R vdash P wedge R$
I can do box proof like:
$dfrac{dfrac{Q wedge P^{~text{(assumption)}}}{P}{^text{($wedge$-elimination)}}quad R^{~text{(assumption)}}}{Pwedge R}{^text{($wedge$ introduction)}}
\text{ (Q.E.D.)}$
But what about when I'm asked to prove $A rightarrow (B rightarrow A)$?
Do I just start with assumption as $A$?
logic natural-deduction
logic natural-deduction
asked Nov 13 at 21:17
Ansar Al
414
asked Nov 13 at 21:17
Ansar Al
414
edited Nov 14 at 12:56
asked Nov 13 at 21:17
Ansar Al
414
asked Nov 13 at 21:17
Ansar Al
414
asked Nov 13 at 21:17
Ansar Al
414
414
3
You use the $to$ introduction rule, which means you start with a temporary assumption of $A$ and derive $Bto A$ (also using $to$-introduction).
– spaceisdarkgreen
Nov 13 at 21:39
add a comment |
3
You use the $to$ introduction rule, which means you start with a temporary assumption of $A$ and derive $Bto A$ (also using $to$-introduction).
– spaceisdarkgreen
Nov 13 at 21:39
3
3
You use the $to$ introduction rule, which means you start with a temporary assumption of $A$ and derive $Bto A$ (also using $to$-introduction).
– spaceisdarkgreen
Nov 13 at 21:39
You use the $to$ introduction rule, which means you start with a temporary assumption of $A$ and derive $Bto A$ (also using $to$-introduction).
– spaceisdarkgreen
Nov 13 at 21:39
add a comment |
1 Answer
1
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votes
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2
down vote
But what about when I'm asked to prove $A→(B→A)$?
Do I just use start with assumption as $A$?
Yes, assume $A$. Next assume $B$, and lo, somehow derive $A$ from those assumptions. Finally use conditional introduction a few times to discharge those assumptions.
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
But what about when I'm asked to prove $A→(B→A)$?
Do I just use start with assumption as $A$?
Yes, assume $A$. Next assume $B$, and lo, somehow derive $A$ from those assumptions. Finally use conditional introduction a few times to discharge those assumptions.
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
add a comment |
up vote
2
down vote
But what about when I'm asked to prove $A→(B→A)$?
Do I just use start with assumption as $A$?
Yes, assume $A$. Next assume $B$, and lo, somehow derive $A$ from those assumptions. Finally use conditional introduction a few times to discharge those assumptions.
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
add a comment |
up vote
2
down vote
up vote
2
down vote
But what about when I'm asked to prove $A→(B→A)$?
Do I just use start with assumption as $A$?
Yes, assume $A$. Next assume $B$, and lo, somehow derive $A$ from those assumptions. Finally use conditional introduction a few times to discharge those assumptions.
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
But what about when I'm asked to prove $A→(B→A)$?
Do I just use start with assumption as $A$?
Yes, assume $A$. Next assume $B$, and lo, somehow derive $A$ from those assumptions. Finally use conditional introduction a few times to discharge those assumptions.
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
edited Nov 13 at 23:12
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
answered Nov 13 at 23:03
Graham Kemp
83.9k43378
83.9k43378
add a comment |
add a comment |
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You use the $to$ introduction rule, which means you start with a temporary assumption of $A$ and derive $Bto A$ (also using $to$-introduction).
– spaceisdarkgreen
Nov 13 at 21:39