ElistheFox
New member
- Joined
- Mar 8, 2021
- Messages
- 4
Dear community,
since it is my first time solving this type of logic, I would like to share it with you - any feedback will be very much appreciated!
This is the first line: ((φ⇒ψ)∧φ)⇒ψ
My solution:
Val((φ⇒ψ)∧φ)⇒ψ =? 1
Val(φ)=a ; val(ψ) = b ; a,b e (0,1)
= lt (val((φ⇒ψ) ; val (∧φ)⇒ψ)
= lt(lt(val(φ), val(ψ), max (val(φ)), val(ψ)))
= lt(lt(val(φ), val(ψ), max (lt(val(φ), 0), val(ψ)))
=lt(lt(a,b), max (lt(a,0),b)
a <= b lt(1, max(0,b)
lt (1,b)
= b != 1
a > b lt(b, max, (0,b))
lt (b,b)
=b != 1
Not a 1-tautology for Gödel logic.
Thank you for your time!
since it is my first time solving this type of logic, I would like to share it with you - any feedback will be very much appreciated!
This is the first line: ((φ⇒ψ)∧φ)⇒ψ
My solution:
Val((φ⇒ψ)∧φ)⇒ψ =? 1
Val(φ)=a ; val(ψ) = b ; a,b e (0,1)
= lt (val((φ⇒ψ) ; val (∧φ)⇒ψ)
= lt(lt(val(φ), val(ψ), max (val(φ)), val(ψ)))
= lt(lt(val(φ), val(ψ), max (lt(val(φ), 0), val(ψ)))
=lt(lt(a,b), max (lt(a,0),b)
a <= b lt(1, max(0,b)
lt (1,b)
= b != 1
a > b lt(b, max, (0,b))
lt (b,b)
=b != 1
Not a 1-tautology for Gödel logic.
Thank you for your time!