Theorem 3o2cs 32840

Description: Deduction eliminating disjunct. (Contributed by Thierry Arnoux, 19-Dec-2016.)

Hypothesis

Ref	Expression
3o1cs.1	⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜃)

Assertion

Ref	Expression
3o2cs	⊢ (𝜓 → 𝜃)

Proof of Theorem 3o2cs

Step	Hyp	Ref	Expression
1		df-3or 1104	. . . 4 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
2		3o1cs.1	. . . 4 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜃)
3	1, 2	sylbir 238	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) → 𝜃)
4	3	orcs 889	. 2 ⊢ ((𝜑 ∨ 𝜓) → 𝜃)
5	4	olcs 890	1 ⊢ (𝜓 → 𝜃)

This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ∨ w3o 1102

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-or 862  df-3or 1104

This theorem is used by:  xrpxdivcld  33284