You are an expert in program synthesis. You are tasked with solving a Syntax-Guided Synthesis (SyGuS) problem. Your goal is to output a function that should produce outputs that satisfy a series of constraints when given specific inputs.

Question:
(set-logic BV)

(synth-fun inv ((s (BitVec 4)) (t (BitVec 4))) (BitVec 4))

(declare-var s (BitVec 4))
(declare-var t (BitVec 4))
(define-fun udivtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) #xF (bvudiv a b)))
(define-fun uremtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) a (bvurem a b)))
(define-fun min () (BitVec 4)
    (bvnot (bvlshr (bvnot #x0) #x1)))
(define-fun max () (BitVec 4)
    (bvnot min))
(define-fun l ((s (BitVec 4)) (t (BitVec 4))) Bool
    (bvugt (udivtotal s (inv s t)) t))
(define-fun SC ((s (BitVec 4)) (t (BitVec 4))) Bool
    (bvult t (bvnot #x0)))
(constraint (=> (SC s t) (l s t)))

(check-synth)
Solution:
(define-fun inv ((s (BitVec 4)) (t (BitVec 4))) (BitVec 4) #b0000)

Question:
(set-logic BV)

(synth-fun inv ((s (BitVec 4)) (t (BitVec 4))) (BitVec 4))

(declare-var s (BitVec 4))
(declare-var t (BitVec 4))
(define-fun udivtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) #xF (bvudiv a b)))
(define-fun uremtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) a (bvurem a b)))
(define-fun min () (BitVec 4)
    (bvnot (bvlshr (bvnot #x0) #x1)))
(define-fun max () (BitVec 4)
    (bvnot min))
(define-fun l ((s (BitVec 4)) (t (BitVec 4))) Bool
    (bvugt (bvashr s (inv s t)) t))
(define-fun SC ((s (BitVec 4)) (t (BitVec 4))) Bool
    (or (bvslt s (bvlshr s (bvnot t))) (bvult t s)))
(constraint (=> (SC s t) (l s t)))

(check-synth)
Solution:
(define-fun inv ((s (BitVec 4)) (t (BitVec 4))) (BitVec 4) (bvand s #b1000))

Question:
(set-logic BV)

(synth-fun inv ((s (BitVec 4)) (t (BitVec 4))) (BitVec 4))

(declare-var s (BitVec 4))
(declare-var t (BitVec 4))
(define-fun udivtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) #xF (bvudiv a b)))
(define-fun uremtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) a (bvurem a b)))
(define-fun min () (BitVec 4)
    (bvnot (bvlshr (bvnot #x0) #x1)))
(define-fun max () (BitVec 4)
    (bvnot min))
(define-fun l ((s (BitVec 4)) (t (BitVec 4))) Bool
    (bvsle (bvor (inv s t) s) t))
(define-fun SC ((s (BitVec 4)) (t (BitVec 4))) Bool
    (bvsge t (bvor s min)))
(constraint (=> (SC s t) (l s t)))

(check-synth)
Solution:
(define-fun inv ((s (BitVec 4)) (t (BitVec 4))) (BitVec 4) #b1000)

Question:
(set-logic BV)

(synth-fun inv ((s (BitVec 4)) (t (BitVec 4))) (BitVec 4))

(declare-var s (BitVec 4))
(declare-var t (BitVec 4))
(define-fun udivtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) #xF (bvudiv a b)))
(define-fun uremtotal ((a (BitVec 4)) (b (BitVec 4))) (BitVec 4)
    (ite (= b #x0) a (bvurem a b)))
(define-fun min () (BitVec 4)
    (bvnot (bvlshr (bvnot #x0) #x1)))
(define-fun max () (BitVec 4)
    (bvnot min))
(define-fun l ((s (BitVec 4)) (t (BitVec 4))) Bool
    (bvult (bvashr s (inv s t)) t))
(define-fun SC ((s (BitVec 4)) (t (BitVec 4))) Bool
    (and (not (and (not (bvult s t)) (bvslt s #x0))) (not (= t #x0))))
(constraint (=> (SC s t) (l s t)))

(check-synth)
Solution:
