Recent content by stan1992
-
S
MHB Understanding the Logic of Quantified Statements
≡(∀x)(∃y)(∃z)[(F(x,y)∧G(x,z))∧¬H(y,z)]- stan1992
- Post #9
- Forum: Set Theory, Logic, Probability, Statistics
-
S
MHB Understanding the Logic of Quantified Statements
(∀x)(∃y)(∃z)(¬F(x,y)V¬G(x,z)∧¬H(y,z)) ¬G is because of Dem. and ¬H from the Def→ right?- stan1992
- Post #7
- Forum: Set Theory, Logic, Probability, Statistics
-
S
MHB Prove that the expression is a valid argument using the deduction method
Prove that the expression below is a valid argument using the deduction method (that is using equivalences and rules of inference in a proof sequence) (∃x)[P(x) → Q(x)]∧(∀y)[Q(y) → R(y)]∧(∀x)P(x) → (∃x)R(x) 1.(∃x)[P(x) → Q(x)] prem 2.(∀y)[Q(y) → R(y)] prem 3.P(x)→Q(x) 1,ui 4.P(x) 2,ui...- stan1992
- Post #3
- Forum: Set Theory, Logic, Probability, Statistics
-
S
MHB Understanding the Logic of Quantified Statements
≡(∀x)(∃y)(∃z)[¬F(x,y)VG(x,z))∧H(y,z)] Would this be it?- stan1992
- Post #5
- Forum: Set Theory, Logic, Probability, Statistics
-
S
MHB Understanding the Logic of Quantified Statements
≡(∀x)(∃y)(∃z)[¬(F(x,y)∧G(x,z))→¬H(y,z)] ≡(∀x)(∃y)(∃z)[¬F(x,y)∨¬G(x,z)→¬H(y,z)] Is this correct?- stan1992
- Post #3
- Forum: Set Theory, Logic, Probability, Statistics
-
S
MHB Prove that the expression is a valid argument using the deduction method
(∃x)[P(x) → Q(x)]∧(∀y)[Q(y) → R(y)]∧(∀x)P(x) → (∃x)R(x)- stan1992
- Thread
- Argument Expression Method
- Replies: 3
- Forum: Set Theory, Logic, Probability, Statistics
-
S
MHB Understanding the Logic of Quantified Statements
∃x∀y∀z[(F(x, y)∧G(x,z)) → H(y,z)]- stan1992
- Thread
- Logic
- Replies: 9
- Forum: Set Theory, Logic, Probability, Statistics