- #1
poutsos.A
- 102
- 1
[tex]\forall a\forall b[/tex][( a>0 & b>0)------> (a[tex]\leq b[/tex] <------>[tex]a^{2}[/tex][tex]\leq b^{2}[/tex])].
or in words: for all a and for all b , if a>0 and b>0 then .a[tex]\leq b[/tex] iff [tex]a^{2}[/tex][tex]\leq b^{2}[/tex]
is there a possibility for a proof within the predicate calculus??
or in words: for all a and for all b , if a>0 and b>0 then .a[tex]\leq b[/tex] iff [tex]a^{2}[/tex][tex]\leq b^{2}[/tex]
is there a possibility for a proof within the predicate calculus??
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