Propositional Logic - Deduction
$begingroup$
So i have to prove that: $${neg Ato B,Ato C,Bto D}vdash neg Cto D$$
I can use logical axioms, modus ponens and 'metatheorems'.
Logical axioms:
- φ→(ψ→φ)
- (φ→(ψ→χ))→((φ→ψ)→(φ→χ))
- (¬φ→¬ψ)→(ψ→φ)
Also i can use modus ponens(the only rule i can use) and metatheorems
Some thoughts:So i started experimenting with all $3$ tools i have, started asking myself is any of the hypotheses can give as something new using the logic axioms but then i stalled, and modus ponens can't do much on it's own knowing these hypotheses atleast.My next thought was that i have to use those 2 metatheorems in order to actualy prove one part of $neg Cto D$ (based on metatheorem 2) meaning i use as a hypothesis $neg C$ to prove $D$ but i am stuck and i don't undestand even how to start.
logic propositional-calculus
edited Dec 14 '18 at 10:34
drhab
98.7k544129
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
$endgroup$
add a comment |
$begingroup$
So i have to prove that: $${neg Ato B,Ato C,Bto D}vdash neg Cto D$$
I can use logical axioms, modus ponens and 'metatheorems'.
Logical axioms:
- φ→(ψ→φ)
- (φ→(ψ→χ))→((φ→ψ)→(φ→χ))
- (¬φ→¬ψ)→(ψ→φ)
Also i can use modus ponens(the only rule i can use) and metatheorems
Some thoughts:So i started experimenting with all $3$ tools i have, started asking myself is any of the hypotheses can give as something new using the logic axioms but then i stalled, and modus ponens can't do much on it's own knowing these hypotheses atleast.My next thought was that i have to use those 2 metatheorems in order to actualy prove one part of $neg Cto D$ (based on metatheorem 2) meaning i use as a hypothesis $neg C$ to prove $D$ but i am stuck and i don't undestand even how to start.
logic propositional-calculus
edited Dec 14 '18 at 10:34
drhab
98.7k544129
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
$endgroup$
add a comment |
$begingroup$
So i have to prove that: $${neg Ato B,Ato C,Bto D}vdash neg Cto D$$
I can use logical axioms, modus ponens and 'metatheorems'.
Logical axioms:
- φ→(ψ→φ)
- (φ→(ψ→χ))→((φ→ψ)→(φ→χ))
- (¬φ→¬ψ)→(ψ→φ)
Also i can use modus ponens(the only rule i can use) and metatheorems
Some thoughts:So i started experimenting with all $3$ tools i have, started asking myself is any of the hypotheses can give as something new using the logic axioms but then i stalled, and modus ponens can't do much on it's own knowing these hypotheses atleast.My next thought was that i have to use those 2 metatheorems in order to actualy prove one part of $neg Cto D$ (based on metatheorem 2) meaning i use as a hypothesis $neg C$ to prove $D$ but i am stuck and i don't undestand even how to start.
logic propositional-calculus
edited Dec 14 '18 at 10:34
drhab
98.7k544129
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
$endgroup$
So i have to prove that: $${neg Ato B,Ato C,Bto D}vdash neg Cto D$$
I can use logical axioms, modus ponens and 'metatheorems'.
Logical axioms:
- φ→(ψ→φ)
- (φ→(ψ→χ))→((φ→ψ)→(φ→χ))
- (¬φ→¬ψ)→(ψ→φ)
Also i can use modus ponens(the only rule i can use) and metatheorems
Some thoughts:So i started experimenting with all $3$ tools i have, started asking myself is any of the hypotheses can give as something new using the logic axioms but then i stalled, and modus ponens can't do much on it's own knowing these hypotheses atleast.My next thought was that i have to use those 2 metatheorems in order to actualy prove one part of $neg Cto D$ (based on metatheorem 2) meaning i use as a hypothesis $neg C$ to prove $D$ but i am stuck and i don't undestand even how to start.
logic propositional-calculus
logic propositional-calculus
edited Dec 14 '18 at 10:34
drhab
98.7k544129
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
edited Dec 14 '18 at 10:34
drhab
98.7k544129
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
edited Dec 14 '18 at 10:34
drhab
98.7k544129
edited Dec 14 '18 at 10:34
drhab
98.7k544129
edited Dec 14 '18 at 10:34
drhab
98.7k544129
98.7k544129
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
asked Dec 14 '18 at 10:28
AgaeusAgaeus
626
626
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
1) Using Modus Ponens and your Metatheorem 1, prove Hypothetical Syllogism :
$varphi to psi, psi to chi vdash varphi to chi$.
2) Using axioms, prove Contraposition :
$(varphi to psi) vdash (lnot psi to lnot varphi)$.
Finally, use them to derive :
$lnot C to B, B to D vdash lnot C to D$.
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
$begingroup$
Thank You, i add the solved problem so anyone can benefit.
$endgroup$
– Agaeus
Dec 14 '18 at 11:23
add a comment |
$begingroup$
From contraposition have A->C to ~C->~A and from metateorem 1 my problem becomes
{~A→B,~C→~A,B→D,~C}⊢ D
- ~C
2.~C->~A - ~A MP 2,1
4.~A->B - B MP 4,3
6.B->D - D MP 5,6
Which gives us that our hypothesis ~C is correct and the original is proven.
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
1) Using Modus Ponens and your Metatheorem 1, prove Hypothetical Syllogism :
$varphi to psi, psi to chi vdash varphi to chi$.
2) Using axioms, prove Contraposition :
$(varphi to psi) vdash (lnot psi to lnot varphi)$.
Finally, use them to derive :
$lnot C to B, B to D vdash lnot C to D$.
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
$begingroup$
Thank You, i add the solved problem so anyone can benefit.
$endgroup$
– Agaeus
Dec 14 '18 at 11:23
add a comment |
$begingroup$
Hint
1) Using Modus Ponens and your Metatheorem 1, prove Hypothetical Syllogism :
$varphi to psi, psi to chi vdash varphi to chi$.
2) Using axioms, prove Contraposition :
$(varphi to psi) vdash (lnot psi to lnot varphi)$.
Finally, use them to derive :
$lnot C to B, B to D vdash lnot C to D$.
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
$begingroup$
Thank You, i add the solved problem so anyone can benefit.
$endgroup$
– Agaeus
Dec 14 '18 at 11:23
add a comment |
$begingroup$
Hint
1) Using Modus Ponens and your Metatheorem 1, prove Hypothetical Syllogism :
$varphi to psi, psi to chi vdash varphi to chi$.
2) Using axioms, prove Contraposition :
$(varphi to psi) vdash (lnot psi to lnot varphi)$.
Finally, use them to derive :
$lnot C to B, B to D vdash lnot C to D$.
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
Hint
1) Using Modus Ponens and your Metatheorem 1, prove Hypothetical Syllogism :
$varphi to psi, psi to chi vdash varphi to chi$.
2) Using axioms, prove Contraposition :
$(varphi to psi) vdash (lnot psi to lnot varphi)$.
Finally, use them to derive :
$lnot C to B, B to D vdash lnot C to D$.
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
answered Dec 14 '18 at 10:46
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
64.7k448112
$begingroup$
Thank You, i add the solved problem so anyone can benefit.
$endgroup$
– Agaeus
Dec 14 '18 at 11:23
add a comment |
$begingroup$
Thank You, i add the solved problem so anyone can benefit.
$endgroup$
– Agaeus
Dec 14 '18 at 11:23
$begingroup$
Thank You, i add the solved problem so anyone can benefit.
$endgroup$
– Agaeus
Dec 14 '18 at 11:23
$begingroup$
Thank You, i add the solved problem so anyone can benefit.
$endgroup$
– Agaeus
Dec 14 '18 at 11:23
add a comment |
$begingroup$
From contraposition have A->C to ~C->~A and from metateorem 1 my problem becomes
{~A→B,~C→~A,B→D,~C}⊢ D
- ~C
2.~C->~A - ~A MP 2,1
4.~A->B - B MP 4,3
6.B->D - D MP 5,6
Which gives us that our hypothesis ~C is correct and the original is proven.
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
$endgroup$
add a comment |
$begingroup$
From contraposition have A->C to ~C->~A and from metateorem 1 my problem becomes
{~A→B,~C→~A,B→D,~C}⊢ D
- ~C
2.~C->~A - ~A MP 2,1
4.~A->B - B MP 4,3
6.B->D - D MP 5,6
Which gives us that our hypothesis ~C is correct and the original is proven.
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
$endgroup$
add a comment |
$begingroup$
From contraposition have A->C to ~C->~A and from metateorem 1 my problem becomes
{~A→B,~C→~A,B→D,~C}⊢ D
- ~C
2.~C->~A - ~A MP 2,1
4.~A->B - B MP 4,3
6.B->D - D MP 5,6
Which gives us that our hypothesis ~C is correct and the original is proven.
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
$endgroup$
From contraposition have A->C to ~C->~A and from metateorem 1 my problem becomes
{~A→B,~C→~A,B→D,~C}⊢ D
- ~C
2.~C->~A - ~A MP 2,1
4.~A->B - B MP 4,3
6.B->D - D MP 5,6
Which gives us that our hypothesis ~C is correct and the original is proven.
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
answered Dec 14 '18 at 14:42
AgaeusAgaeus
626
626
add a comment |
add a comment |
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