Metamath Proof Explorer
Description: Deduce a disjunction from another one. Variation on orim12d .
(Contributed by Thierry Arnoux, 18-May-2025)
|
|
Ref |
Expression |
|
Hypotheses |
orim12da.1 |
⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜃 ) |
|
|
orim12da.2 |
⊢ ( ( 𝜑 ∧ 𝜒 ) → 𝜏 ) |
|
|
orim12da.3 |
⊢ ( 𝜑 → ( 𝜓 ∨ 𝜒 ) ) |
|
Assertion |
orim12da |
⊢ ( 𝜑 → ( 𝜃 ∨ 𝜏 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
orim12da.1 |
⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜃 ) |
2 |
|
orim12da.2 |
⊢ ( ( 𝜑 ∧ 𝜒 ) → 𝜏 ) |
3 |
|
orim12da.3 |
⊢ ( 𝜑 → ( 𝜓 ∨ 𝜒 ) ) |
4 |
1
|
ex |
⊢ ( 𝜑 → ( 𝜓 → 𝜃 ) ) |
5 |
2
|
ex |
⊢ ( 𝜑 → ( 𝜒 → 𝜏 ) ) |
6 |
4 5
|
orim12d |
⊢ ( 𝜑 → ( ( 𝜓 ∨ 𝜒 ) → ( 𝜃 ∨ 𝜏 ) ) ) |
7 |
3 6
|
mpd |
⊢ ( 𝜑 → ( 𝜃 ∨ 𝜏 ) ) |