How "with" keyword works in agda ?? and also the code below ?? -

i not able understand clearly. tried learn "with" keyword there have doubt. please !!!

i wanted understand working of "with" , working of code.

something-even : ∀ n → n ⊎ (suc n) something-even 0 = inj₁ 0 something-even (suc n) something-even n ... | inj₁ x = inj₂ (suc-suc x) ... | inj₂ y = inj₁ y (this states either n or successor even). in fact, thm0 can implemented without using recursion!  thm0 : ∀ x → ∃ λ y → y × x less-than y thm0 n something-even n ... | inj₁ x = suc (suc n) , suc-suc x , suc ref ... | inj₂ y = suc n , y , ref 

if familiar haskell, notice in agda there no if-statements , no case.

with pattern-matching of expression result. example, with something-even n evaluates something-even n, , pattern-matches on following lines ... | inj₁ x , ... | inj₂ y. these match expression, see if value constructed using inj₁ or inj₂ constructor, value wrapped them can used in right-hand-side expressions: suc (suc n) , suc-suc x , suc ref uses x determined inj₁ x, , suc n , y , ref uses y determined inj₂ y

the actual constructor names come definition of ⊎ - see type of something-even joins types even n , even (suc n). x in inj₁ x corresponds value of type even n given n, , y in inj₂ y corresponds value of type even (suc n) same n.