Question

P,Q, and R are partitions of of a set.

If P is a refinement of Q and Q is a refinement of R, then P is a refinement of R. (Transitivity).

Prove the above statement.

Answer #1

This the required solution. I hope it will help you. Please give a thumbs up.

suppose f is an integral element function on [0,1], p and q are
partitions of [0,1].
for any partitions A and B, let P = A U B. Then Up-Lp <=
Ua-La.
is this statement true or false? either provide a proof or
counterexample.
cross "element"

suppose f is an integral element function on [0,1], p and q are
partitions of [0,1].
if Up-Lp <= Uq-La, then q belongs to p.
is this statement true or false? either provide a proof or
counterexample.

Prove p ∨ (q ∧ r) ⇒ (p ∨ q) ∧ (p ∨ r) by constructing a proof
tree whose premise is p∨(q∧r) and whose conclusion is
(p∨q)∧(p∨r).

Prove
a)p→q, r→s⊢p∨r→q∨s
b)(p ∨ (q → p)) ∧ q ⊢ p

Prove or disprove that [(p → q) ∧ (p → r)] and [p→ (q ∧ r)] are
logically equivalent.

Prove ((P ∨ ¬Q) ∧ (¬P ∨ R)) → (Q → R)
Hint: this starts with the usual setup for an implication, then
repeatedly uses disjunctive syllogism.

Prove: (p ∧ ¬r → q) and p → (q ∨ r) are biconditional using
natural deduction NOT TRUTH TABLE

Consider the set Pn of all partitions of the set [n] into
non-empty blocks, that is:
Pn = {π | π is a set partition of [n]}
Prove that this is a poset

Using rules of inference prove.
(P -> R) -> ( (Q -> R) -> ((P v Q) -> R) )
Justify each step using rules of inference.

Give the indirect proofs of:
p→q,¬r→¬q,¬r⇒¬p.p→q,¬r→¬q,¬r⇒¬p.
p→¬q,¬r→q,p⇒r.p→¬q,¬r→q,p⇒r.
a∨b,c∧d,a→¬c⇒b.

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