Where is the mistake in this formal proof?

[solakis1]

we have the following formal proof:

i) \(\displaystyle \forall x[\forall y(xy=y)\Longrightarrow x=1]\).......theoren in real Nos

2)\(\displaystyle \forall y(xy=y)\Longrightarrow x=1\)......1,U.E ,x=x

3) \(\displaystyle (x0=0)\Longrightarrow x=1\)...... 2,U.E ,y=0

4)\(\displaystyle \forall A[A.0=0]\)........Theorem in Real Nos5) )\(\displaystyle [x.0=0]\)........4,U.E, A=x

6) x=1........3,5 M.Ponens

7) )\(\displaystyle \forall A[A=1]\)........ 6,U.I

U.E=Universal Elimination

U.I = Universal Introduction

I am afraid to say i find no mistake

 

[Deveno]

The very first line is wrong, it should be:

1) $\forall x[\forall (y \neq 0): (xy = y ) \implies x = 1]$

One can see that your statement is clearly untrue by letting:

$x = 2, y = 0$

so it is not a theorem of the real numbers.

(Moral of the story: don't go proving stupid things by dividing by 0).

 

[solakis1]

The very first line is wrong, it should be:

1) $\forall x[\forall (y \neq 0): (xy = y ) \implies x = 1]$

One can see that your statement is clearly untrue by letting:

$x = 2, y = 0$

so it is not a theorem of the real numbers.

(Moral of the story: don't go proving stupid things by dividing by 0).

1)\(\displaystyle \forall y(xy=y)\).......hypothesis

2)\(\displaystyle (x1=1)\).........1,U.E ,Y=1

3)\(\displaystyle \forall A[A.1=A]\).......Axiom in Real Nos

4))\(\displaystyle [x1=x]\).......3,U.E,A=x

5) x=1........Substituting ( 4) into (2)

6)\(\displaystyle \forall y(xy=y)\Longrightarrow x=1\)....... From (1) to (5) by using the rule of conditional proof

7)\(\displaystyle \forall x[\forall y(xy=y)\Longrightarrow x=1]\)..........6,U.I

Where is the mistake