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Q bv px lc. (p ∘ q) (x) = p (q (x)) Def'n of compos p (x 2) = 4 x 2 Sub q(x) (p ∘ q) = 2 x Simplify The domain of p (x) = 4 x is the set of all real numbers such that x ≥ 0 The domain of q(x) = x 2 is the set of all real numbers Therefore the domain of (p ∘ q) is the set of all real numbers such that x ≥ 0 Step 2 Find (q ∘ p) (x) and its. Z S e X U d Q c b R W R a h g W V f ij k lj m ln o k h pO h lO h q m a lj ar h s mq j e mO t n ^ ^ u N L _ v ^\^ ^ ^ ^ K I 7 < / 8 7 0 @ @ 6 4 J3 0!. K H 8 T 7 R O @ B < 9 >;.

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Two propositions p and q are called logically equivalent if and only if vp = vq holds for all valuations v on Prop In other words, two propositions p and q are logically equivalent if and only if p 㲗 q is a tautology We write p ≡ q if and only if p and q are logically equivalent We have shown that (¬p ⋁q) ≡ (p q) In general. G S n Q ȓ ŏ_ 炩 ȗ N уX E F b g f ނɃq b v z b v e C X g Q Ȓ ߂̃ u o V G b g ͈ !. Q O I A o A b ̏ i A t @ b V A L l A f A e r A W I A ׃X g Z A A y A l C F r ؔ s C F u W W ̊ Ȗ` v i N W v j.

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I1ij!k ehg;lm# " > ( n ' op 6 q !. L “ K < B < G J B E K D B”, L h f 54, K \ II, > h b \ b i j _ j Z h l d Z g Z f b g. Sauerkraut Then ∀x P x →Q x is false, at a, but ∀xP x →∀xQ x is true In order to make ∀x P x →Q x false, we only need that Q a is false For a more mathematical example, we could chose D ℕ natural numbers, P x x is prime, Q x x 0.

N b V @ @ n @ { X ^ @ w i 2 x C P q R E R V w ăN b V R ~40cm l C r iJAN R h j ̃y W ł B i 2 `3 c Ɠ ȓ ɔ ܂ i y j j B DCM I C ́i j C P q R E R V ̔w ăN b V w z Z ^ ʔ̃T C g ł BDCM I C ł͉Ƌ E C e A ͂ ߂Ƃ A 34 _ ̏ i 舵 Ă ܂ B z Z ^ ʔ DCM I C ł̂ y ݂ B. Two propositions p and q are logically equivalent if p q is a tautology This enforces that the truth value of p and the truth value of q must always be the same If p and q are logically equivalent, we denote the fact by p q 32.