How to prove by equational proof
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
rules same as here
I proved via hilbert style to prove
⊢(∀x) A ≡ (∀z) A [x:=z] where z is fresh for A
but struggling with equational proof for
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
logic predicate-logic
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
asked Nov 20 at 18:10
TaeCoda
15
add a comment |
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
rules same as here
I proved via hilbert style to prove
⊢(∀x) A ≡ (∀z) A [x:=z] where z is fresh for A
but struggling with equational proof for
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
logic predicate-logic
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
asked Nov 20 at 18:10
TaeCoda
15
Maybe you have to use the fact/definition : $(∃x)A equiv lnot (∀x) lnot A$.
– Mauro ALLEGRANZA
Nov 21 at 8:02
using that definition, I will get ¬(∀x) ¬A ≡ ¬(∀z) ¬A [x:=z] ? How do you do this equational proof to prove that?
– TaeCoda
Nov 22 at 17:53
add a comment |
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
rules same as here
I proved via hilbert style to prove
⊢(∀x) A ≡ (∀z) A [x:=z] where z is fresh for A
but struggling with equational proof for
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
logic predicate-logic
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
asked Nov 20 at 18:10
TaeCoda
15
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
rules same as here
I proved via hilbert style to prove
⊢(∀x) A ≡ (∀z) A [x:=z] where z is fresh for A
but struggling with equational proof for
⊢(∃x)A≡(∃z)A[x:=z] where z is fresh for A
logic predicate-logic
logic predicate-logic
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
asked Nov 20 at 18:10
TaeCoda
15
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
asked Nov 20 at 18:10
TaeCoda
15
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
edited Nov 21 at 8:03
Mauro ALLEGRANZA
64.2k448112
64.2k448112
asked Nov 20 at 18:10
TaeCoda
15
asked Nov 20 at 18:10
TaeCoda
15
asked Nov 20 at 18:10
TaeCoda
15
15
Maybe you have to use the fact/definition : $(∃x)A equiv lnot (∀x) lnot A$.
– Mauro ALLEGRANZA
Nov 21 at 8:02
using that definition, I will get ¬(∀x) ¬A ≡ ¬(∀z) ¬A [x:=z] ? How do you do this equational proof to prove that?
– TaeCoda
Nov 22 at 17:53
add a comment |
Maybe you have to use the fact/definition : $(∃x)A equiv lnot (∀x) lnot A$.
– Mauro ALLEGRANZA
Nov 21 at 8:02
using that definition, I will get ¬(∀x) ¬A ≡ ¬(∀z) ¬A [x:=z] ? How do you do this equational proof to prove that?
– TaeCoda
Nov 22 at 17:53
Maybe you have to use the fact/definition : $(∃x)A equiv lnot (∀x) lnot A$.
– Mauro ALLEGRANZA
Nov 21 at 8:02
Maybe you have to use the fact/definition : $(∃x)A equiv lnot (∀x) lnot A$.
– Mauro ALLEGRANZA
Nov 21 at 8:02
using that definition, I will get ¬(∀x) ¬A ≡ ¬(∀z) ¬A [x:=z] ? How do you do this equational proof to prove that?
– TaeCoda
Nov 22 at 17:53
using that definition, I will get ¬(∀x) ¬A ≡ ¬(∀z) ¬A [x:=z] ? How do you do this equational proof to prove that?
– TaeCoda
Nov 22 at 17:53
add a comment |
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Maybe you have to use the fact/definition : $(∃x)A equiv lnot (∀x) lnot A$.
– Mauro ALLEGRANZA
Nov 21 at 8:02
using that definition, I will get ¬(∀x) ¬A ≡ ¬(∀z) ¬A [x:=z] ? How do you do this equational proof to prove that?
– TaeCoda
Nov 22 at 17:53