Theorem orim12da 980

Description: Deduce a disjunction from another one. Variation on orim12d 979. (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 418	. . 3 ⊢ (𝜑 → (𝜓 → 𝜃))
4		orim12da.2	. . . 4 ⊢ ((𝜑 ∧ 𝜒) → 𝜏)
5	4	ex 418	. . 3 ⊢ (𝜑 → (𝜒 → 𝜏))
6	3, 5	orim12d 979	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜒) → (𝜃 ∨ 𝜏)))
7	1, 6	mpd 16	1 ⊢ (𝜑 → (𝜃 ∨ 𝜏))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∨ wo 861

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-an 402  df-or 862

This theorem is used by:  lnincplng  29103  plngmiropp  29113  prlngsym  29228  ricdomn1  33640  drngmxidlr  33791  drnglring  33813  dflring2  33814  dflringlem2  33816  dflring3  33818  rsprprmprmidl  33843  rsprprmprmidlb  33844  rprmirredb  33853  rprmdvdsprod  33855  mplidomlem  33948  rtelextdg2  34148