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  29295  plngmiropp  29305  prlngsym  29452  ricdomn1  33926  drngmxidlr  34086  drnglring  34108  dflring2  34109  dflringlem2  34111  dflring3  34113  rsprprmprmidl  34138  rsprprmprmidlb  34139  rprmirredb  34148  rprmdvdsprod  34150  mplidomlem  34242  rtelextdg2  34442  squeezedltsq  48127  inlinecirc02plem  50113  parmapcl2  50985