Metamath Proof Explorer


Theorem 3jca

Description: Join consequents with conjunction. (Contributed by NM, 9-Apr-1994)

Ref Expression
Hypotheses 3jca.1 ( 𝜑𝜓 )
3jca.2 ( 𝜑𝜒 )
3jca.3 ( 𝜑𝜃 )
Assertion 3jca ( 𝜑 → ( 𝜓𝜒𝜃 ) )

Proof

Step Hyp Ref Expression
1 3jca.1 ( 𝜑𝜓 )
2 3jca.2 ( 𝜑𝜒 )
3 3jca.3 ( 𝜑𝜃 )
4 1 2 3 jca31 ( 𝜑 → ( ( 𝜓𝜒 ) ∧ 𝜃 ) )
5 df-3an ( ( 𝜓𝜒𝜃 ) ↔ ( ( 𝜓𝜒 ) ∧ 𝜃 ) )
6 4 5 sylibr ( 𝜑 → ( 𝜓𝜒𝜃 ) )