To the following rigorous proof examined in another thread i was asked to show the way that the laws of logic are involved.
1) (-1)x = (-1)x + 0................................................. .................................................. ..............by using the axiom : for all ,a: a + 0 = a
2) (-1)x + 0 = (-1)x + ( x + (-x))............................................... ................................................by using the axiom: for all ,a : a + (-a) = 0
3) (-1)x + (x + (-x)) = ((-1)x + x) + (-x)................................................ ....................................by using the axiom : for all a,b.c : a + ( b + c) = ( a + b) + c
4) ((-1)x + x) + (-x) = ( x + (-1)x) + (-x)................................................ ....................................by using the axiom : for all a,b : a+ b = b + a
5) ( x + (-1)x) + (-x) = ( 1x + (-1)x) + (-x)................................................ .................................by using the axiom: for all ,a : 1a = a
6) ( 1 + (-1))x + (-x) = ( 1x + (-1)x) + (-x)................................................ .................................by using the axiom : for all a,b,c : (a + b)c = ac + bc
7) ( 1 + (-1))x + (-x) = 0x + (-x)................................................ ...............................................by using the axiom : for all ,a : a + (-a) = 0
8) 0x + (-x) = 0 + (-x)................................................ .................................................. ...........by using the theorem : for all, a: 0a = 0
9) 0 + (-x) = (-x) + 0................................................. .................................................. ............by using the axiom : for all a,b : a + b = b + a
10) (-x) + 0 = -x................................................. .................................................. ..................by using the axiom : for all ,a : a + 0 = a
11) HENCE..FOR ALL ,X : (-1)X= -X....................................
For line (1) the axiom used is: for all ,a : a + 0 = a . And by the use of Universal Elimination where we put a= (-1)x we obtain:
1) (-1)x + 0 = (-1)x.
For line (2) the axiom used is : for all ,a : a + (-a) = 0. And by using Universal Elimination where we put a = x we have :
1a) x + (-x) = 0
And substituting (1a) into (1) we get :
2) (-1)x + ( x + (-x)) = (-1)x
For line (3) the axiom used is : for all a,b,c: a + ( b + c) = ( a + b) + c .And by using Universal Elimination where we put a = (-1)x ,b = x ,c = -x we have:
2a) (-1)x + ( x + (-x)) = ((-1)x + x) + (-x).
And substituting (2a) into (2) we get :
3) ((-1)x + x) + (-x) = (-1)x
For line (4) the axiom used is : for all a,b : a + b = b + a .And by using Universal Elimination where we put a = (-1)x ,b = x we have :
3a) ((-1)x + x) = ( x + (-1)x).
And substituting (3a) into (3) we get:
4) ( x + (-1)x) + (-x) = (-1)x.
For line (5) the axiom used is : for all ,a : 1a = a.And by using Universal Elimination where we put a = x we have:
4a) ( x + (-1)x) = ( 1x + (-1)x).
And substituting (4a) into (4) we get:
5) ( 1x + (-1)x)+ (-x) = (-1)x.
For line (6) the axiom used is: for all a,b,c: ( a + b )c = ac + bc.And by using Universal Elimination where we put a = 1, b = -1 , c = x we have:
5a) ( 1 + (-1))x = ( 1x + (-1)x).
And substituting (5a) into (5) we get:
6)( 1 + (-1))x + (-x) = (-1)x.
For line (7) the axiom used is : for all ,a : a + (-a) = 0. And by using U.E where we put a = 1 we have :
6a) 1 + (-1) =0.
And substituting (6a) into (6) we get:
7) 0x + (-x) = (-1)x.
For line (8) the theorem used is : for all ,a : 0a =0 .And by using U.E where we put a =x we have :
7a) 0x = 0 .
And substituting (7a) into (7) we get :
8) 0 + (-x) = (-1)x.
For line (9) the axiom used is: for all ,a ,b : a + b = b + a .And by using U.E where we put a = 0 , b = (-x) we have:
8a) -x + 0 = 0 + (-x).
And substituting (8a) into (8) we get :
9) -x + 0 = (-1)x .
For line (10) the axiom used is : for all ,a : a + 0 = a .And by using U.E where we put a = -x we have :
9a) -x + 0 = -x .
And substituting (9a) into (9) we get:
10) -x = (-1)x.
Is that correct?? .
Thanks


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