Are these equations equivalent? Please prove the answer either via truth table equivalence.   P1 = A ∨ ( B∧ (C ∨ ¬D )) ∨ (( D ∨ E) ∧ ( E ∨ F ∨ G )) P2= A ∨ ¬ (( ¬B ∨ ¬C ) ∧ ( ¬B∨ D )) ∨ ((( E ∧ D ) ∨ E ) ∨ (F ∧ ( E ∨ D )) ∨ (G ∧ (D ∨ E)))

Part 1: Choose two of the proofs below and use one of the indirect proof techniques (reductio ad absurdum or conditional proof) presented in Chapter 8 to demonstrate the validity of the arguments. The proofs below may use any of the rules of inference or replacement rules given in Chapter 8. 1.(G • P) → K, E → Z, ~P → ~ Z, G → (E v L), therefore, (G • ~L) → K 2.(S v T) ↔ ~E, S → (F • ~G), A → W, T → ~W, therefore, (~E • A) → ~G 3.(S v T) v (U v W), therefore, (U v T) v (S v W) 4.~Q → (L → F), Q → ~A, F → B, L, therefore, ~A v B 5.~S → (F → L), F → (L → P), therefore, ~S → (F → P) Part 2: Below are basic arguments in English. Choose two arguments and translate those argument into the symbolism of predicate logic. You do not need to do a proof. Every fetus has an immortal soul. A thing has an immortal soul only if it has a right to life. Hence, every fetus has a right to life. (Fx = x is a fetus, Sx = x has an immortal soul, Rx = x has a right to life). Some wars are just. No war of…

Use the laws of propositional logic (logical equivalences) to show the following equivalency by choosing to change one, and only one, side of the equivalence expression.  (¬p→q)∨¬q ≡ T

¬P ∧¬Q≡¬(P ∨Q)   Prove the following equivalency relation (Proof, Justification/explanation):

Simplify the following logic formulas. Use a truth table to show that your simplification is correct. (a.) (p∧q)∨q(p∧q)∨q (b.) p→(p→q)p→(p→q) (c.) ¬(¬p∧¬q)¬(¬p∧¬q) "Simplify" means: find the simplest/smallest formula that means the same thing.

Are the following equations equivalent? Please prove the answer via truth table equivalence.P1 = ((A∨¬B) ∧ C )∨¬D P2=((¬A ∧C) ∨ ( B ∧C)) ∧ D

4. For each of the following propositional formulae, use the laws of propositional logic to prove it is either a tautology or contradiction. a.   ((¬p→q)∧(¬p→¬q))→p((¬p→q)∧(¬p→¬q))→p b.   ((p∧q)∨¬(¬p∨r))∧¬p