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The Reasoning Operators ⇒ and ⇔

$rArr$ and $hArr$ are not propositional operators

different syntax:
$(2x-6 = 0 harr 2x = 6) harr x = 3$ is the same as
$2x-6 = 0 harr 2x = 6 harr x = 3$
$(2x-6 = 0 hArr 2x = 6) hArr x = 3$ is a syntax error
verbose_off ok_text You see! modulo 13 (2x-6 = 0 <-> 2x = 6) <-> x = 3 /**/<=> 2x-6 = 0 <-> 2x = 6 <-> x = 3 /**/<=> x=3

verbose_off ok_text You changed something! modulo 13 (2x-6 = 0 <=> 2x = 6) <=> x = 3

verbose_off ok_text You changed something! tree_compare 2x-6 = 0 <-> 2x = 6 <-> x = 3; 2x-6 = 0 <=> 2x = 6 <=> x = 3
so negation cannot be applied to $rArr$ and $hArr$
different meaning of chains:
$2x-6 = 0 hArr 2x = 6 hArr x = 3$ is valid for every $x in RR$
$2x-6 = 0 harr 2x = 6 harr x = 3$ yields F when $x ne 3$
different reaction to context information
case $x < 0$ : … $x^2 = 9 hArr x = -3$

We must — and can — and do — have U

$hArr$ F