Question

Let a(x,y), b(x,y) and c(x,y) be three statements with variables x and y chosen from some universe . Consider the following statement:
$\left(\exists x\right)\left(\forall y\right)[(a(x,y)\wedge (b(x,y))\wedge \rightharpoondown c(x,y)]$
Which one of the following is its equivalent ?

• A. $\rightharpoondown \left(\forall x\right)\left(\exists y\right)[(a(x,y)\vee b(x,y))\to c(x,y)]$
• B. $\left(\forall x\right)\left(\exists y\right)[(a(x,y)\vee b(x,y))\to c(x,y)]$
• C. $\rightharpoondown \left(\forall x\right)\left(\exists y\right)[(a(x,y)\wedge b(x,y))\to c(x,y)]$
• D. $\left(\exists x\right)\left(\forall y\right)[(a(x,y)\vee b(x,y))\wedge \rightharpoondown c(x,y)]$

Answer 1

The correct option is C $\rightharpoondown \left(\forall x\right)\left(\exists y\right)[(a(x,y)\wedge b(x,y))\to c(x,y)]$

Choice (c) is
$\rightharpoondown \left(\forall x\right)\left(\exists y\right)\left[\right(a(x,y)\wedge b(x,y))\to c(x,y\left)\right]$
$\equiv \rightharpoondown \left(\forall x\right)\left(\exists y\right)[a\wedge b\to c]$
$\equiv \rightharpoondown \left(\forall x\right)\left(\exists y\right)\left[\right(ab){}^{\prime }+c]$
$\equiv \exists x\forall y\left[\right(ab){}^{\prime }+c]$'
$\equiv \exists x\forall y\left[abc{}^{\prime }\right]\equiv \exists x\forall y[a\wedge b\wedge \rightharpoondown c]$
which is same as the given expression
$\left(\exists x\right)\left(\forall y\right)\left[\right(a(x,y)\wedge b(x,y))\wedge \rightharpoondown c(x,y\left)\right]$