Theorem orim12da 32439

Description: Deduce a disjunction from another one. Variation on orim12d 966. (Contributed by Thierry Arnoux, 18-May-2025.)

Hypotheses

Ref	Expression
orim12da.1	⊢ ((𝜑 ∧ 𝜓) → 𝜃)
orim12da.2	⊢ ((𝜑 ∧ 𝜒) → 𝜏)
orim12da.3	⊢ (𝜑 → (𝜓 ∨ 𝜒))

Assertion

Ref	Expression
orim12da	⊢ (𝜑 → (𝜃 ∨ 𝜏))

Proof of Theorem orim12da

Step	Hyp	Ref	Expression
1		orim12da.3	. 2 ⊢ (𝜑 → (𝜓 ∨ 𝜒))
2		orim12da.1	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜃)
3	2	ex 412	. . 3 ⊢ (𝜑 → (𝜓 → 𝜃))
4		orim12da.2	. . . 4 ⊢ ((𝜑 ∧ 𝜒) → 𝜏)
5	4	ex 412	. . 3 ⊢ (𝜑 → (𝜒 → 𝜏))
6	3, 5	orim12d 966	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜒) → (𝜃 ∨ 𝜏)))
7	1, 6	mpd 15	1 ⊢ (𝜑 → (𝜃 ∨ 𝜏))

Syntax hints:   → wi 4   ∧ wa 395   ∨ wo 847

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 207  df-an 396  df-or 848

This theorem is referenced by:  drngmxidlr  33493  rsprprmprmidl  33537  rsprprmprmidlb  33538  rprmirredb  33547  rprmdvdsprod  33549  rtelextdg2  33761