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| Mirrors > Home > NFE Home > Th. List > mpjaod | GIF version | ||
| Description: Eliminate a disjunction in a deduction. (Contributed by Mario Carneiro, 29-May-2016.) |
| Ref | Expression |
|---|---|
| jaod.1 | ⊢ (φ → (ψ → χ)) |
| jaod.2 | ⊢ (φ → (θ → χ)) |
| jaod.3 | ⊢ (φ → (ψ ∨ θ)) |
| Ref | Expression |
|---|---|
| mpjaod | ⊢ (φ → χ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | jaod.3 | . 2 ⊢ (φ → (ψ ∨ θ)) | |
| 2 | jaod.1 | . . 3 ⊢ (φ → (ψ → χ)) | |
| 3 | jaod.2 | . . 3 ⊢ (φ → (θ → χ)) | |
| 4 | 2, 3 | jaod 369 | . 2 ⊢ (φ → ((ψ ∨ θ) → χ)) |
| 5 | 1, 4 | mpd 14 | 1 ⊢ (φ → χ) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ∨ wo 357 |
| 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 177 df-or 359 |
| This theorem is referenced by: ecase2d 906 fnfreclem3 6320 |
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