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  29267  plngmiropp  29277  prlngsym  29424  ricdomn1  33896  drngmxidlr  34048  drnglring  34070  dflring2  34071  dflringlem2  34073  dflring3  34075  rsprprmprmidl  34100  rsprprmprmidlb  34101  rprmirredb  34110  rprmdvdsprod  34112  mplidomlem  34205  rtelextdg2  34405  squeezedltsq  48005  inlinecirc02plem  49991  parmapcl2  50863