- #1
A.Dasher
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I have this challenge to do but can't seem to be able to understand it completely! I also have test right now and can't concentrate. Please help this poor soul! It will be appreciated!
The first question is:Let F ∈ PROP. Let (h)F be defined by the number of parentheses in the formula F.
a): Give a recursive definition of the function h: PROP -> N (natural numbers) for the number of parentheses in the formula: F= ¬(p1 v (¬p2->p1)) or in this case h(F)=4; Feel free to change the formula if you'd like.
b): Prove with structural induction, using the definition you gave in question a and the function props, that for all proposition formulas F∈ PROP:
1/2 * h(F)+1= props (F);
And the second one is a tree method:
Let A,B,C ∈PROP. Check if these meta-allegations are true or false. Use the tree-method to show if there are counterexamples for these allegations. If there are counterexamples give at least 1 valuation.
a): A->B, ¬B->C |=A->C;
b): A<->(¬B v ¬C), ¬(A -> ¬ C) |= ¬B
Thank you before hand! I really, REALLY appreciate it!
The first question is:Let F ∈ PROP. Let (h)F be defined by the number of parentheses in the formula F.
a): Give a recursive definition of the function h: PROP -> N (natural numbers) for the number of parentheses in the formula: F= ¬(p1 v (¬p2->p1)) or in this case h(F)=4; Feel free to change the formula if you'd like.
b): Prove with structural induction, using the definition you gave in question a and the function props, that for all proposition formulas F∈ PROP:
1/2 * h(F)+1= props (F);
And the second one is a tree method:
Let A,B,C ∈PROP. Check if these meta-allegations are true or false. Use the tree-method to show if there are counterexamples for these allegations. If there are counterexamples give at least 1 valuation.
a): A->B, ¬B->C |=A->C;
b): A<->(¬B v ¬C), ¬(A -> ¬ C) |= ¬B
Thank you before hand! I really, REALLY appreciate it!