the fact that if the trireme is autographic but it is not hydric that the reseau is not a piggyback is correct is wrong.
sent1: that the trireme is autographic thing that is not hydric does not hold if the charge does not eliminate sforzando. sent2: if the charge eliminates sforzando the reseau is not a piggyback. sent3: if the reseau does eliminate sforzando the trireme eliminate sforzando. sent4: something that eliminates sforzando is autographic. sent5: if the trireme is autographic then the fact that the charge is not autographic is not false.
void -> assump1: Let's assume that the fact that the charge does not eliminate sforzando is correct.; sent1 & assump1 -> int1: that the trireme is autographic and is not hydric is not right.; void -> assump2: Let's assume that the trireme is autographic but it is not hydric.; int1 & assump2 -> int2: this is contradiction.; [assump1] & int2 -> int3: the charge eliminates sforzando.; int3 & sent2 -> int4: the reseau is not a kind of a piggyback.; [assump2] & int4 -> hypothesis;
the charge is autographic.
sent4 -> int5: if the trireme does eliminate sforzando then it is autographic.;
$hypothesis$ = the fact that if the trireme is autographic but it is not hydric that the reseau is not a piggyback is correct is wrong. ; $context$ = sent1: that the trireme is autographic thing that is not hydric does not hold if the charge does not eliminate sforzando. sent2: if the charge eliminates sforzando the reseau is not a piggyback. sent3: if the reseau does eliminate sforzando the trireme eliminate sforzando. sent4: something that eliminates sforzando is autographic. sent5: if the trireme is autographic then the fact that the charge is not autographic is not false. ; $proof$ = 
void -> assump1: Let's assume that the fact that the charge does not eliminate sforzando is correct.; sent1 & assump1 -> int1: that the trireme is autographic and is not hydric is not right.; void -> assump2: Let's assume that the trireme is autographic but it is not hydric.; int1 & assump2 -> int2: this is contradiction.; [assump1] & int2 -> int3: the charge eliminates sforzando.; int3 & sent2 -> int4: the reseau is not a kind of a piggyback.; [assump2] & int4 -> hypothesis; __DISPROVED__