Logic.

[fleet light]

[(p↔q)∧(q↔r)∧(r↔p)]⇔[(p→q)∧(q r
→ )∧(r p
→ )]
prove it without truth table

[abstract ice]

[(p↔q)∧(q↔r)∧(r↔p)]⇔[(p→q)∧(q r→ )∧(r p→ )]

in the second part i think it got mispasted or smth

like r p ->

u mean r -> p?

but without truth table i would divide the problem into 2 terms, divided by the big <->, first go from first term to second via ->, then second to first via ->

try to use cool properties