Metamath Proof Explorer


Theorem 3anasss

Description: Associative law for conjunction applied to antecedent (eliminates syllogism). Converse of 3anassrs . (Contributed by Thierry Arnoux, 5-Jul-2026)

Ref Expression
Hypothesis 3anasss.1
|- ( ( ( ( ph /\ ps ) /\ ch ) /\ th ) -> ta )
Assertion 3anasss
|- ( ( ph /\ ( ps /\ ch /\ th ) ) -> ta )

Proof

Step Hyp Ref Expression
1 3anasss.1
 |-  ( ( ( ( ph /\ ps ) /\ ch ) /\ th ) -> ta )
2 13an22anass
 |-  ( ( ph /\ ( ps /\ ch /\ th ) ) <-> ( ( ph /\ ps ) /\ ( ch /\ th ) ) )
3 1 anasss
 |-  ( ( ( ph /\ ps ) /\ ( ch /\ th ) ) -> ta )
4 2 3 sylbi
 |-  ( ( ph /\ ( ps /\ ch /\ th ) ) -> ta )