Metamath Proof Explorer


Theorem mpsyl4anc

Description: An elimination deduction. (Contributed by Alan Sare, 17-Oct-2017)

Ref Expression
Hypotheses mpsyl4anc.1 𝜑
mpsyl4anc.2 𝜓
mpsyl4anc.3 𝜒
mpsyl4anc.4 ( 𝜃𝜏 )
mpsyl4anc.5 ( ( ( ( 𝜑𝜓 ) ∧ 𝜒 ) ∧ 𝜏 ) → 𝜂 )
Assertion mpsyl4anc ( 𝜃𝜂 )

Proof

Step Hyp Ref Expression
1 mpsyl4anc.1 𝜑
2 mpsyl4anc.2 𝜓
3 mpsyl4anc.3 𝜒
4 mpsyl4anc.4 ( 𝜃𝜏 )
5 mpsyl4anc.5 ( ( ( ( 𝜑𝜓 ) ∧ 𝜒 ) ∧ 𝜏 ) → 𝜂 )
6 1 a1i ( 𝜃𝜑 )
7 2 a1i ( 𝜃𝜓 )
8 3 a1i ( 𝜃𝜒 )
9 6 7 8 4 5 syl1111anc ( 𝜃𝜂 )