Theorem 3o1cs 30713

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

Hypothesis

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

Assertion

Ref	Expression
3o1cs	⊢ (𝜑 → 𝜃)

Proof of Theorem 3o1cs

Step	Hyp	Ref	Expression
1		df-3or 1086	. . . 4 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
2		3o1cs.1	. . . 4 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜃)
3	1, 2	sylbir 234	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) → 𝜃)
4	3	orcs 871	. 2 ⊢ ((𝜑 ∨ 𝜓) → 𝜃)
5	4	orcs 871	1 ⊢ (𝜑 → 𝜃)

Syntax hints:   → wi 4   ∨ wo 843   ∨ w3o 1084

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 206  df-or 844  df-3or 1086

This theorem is referenced by:  xrpxdivcld  31111