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Mirrors > Home > NFE Home > Th. List > elif | GIF version |
Description: Membership in a conditional operator. (Contributed by NM, 14-Feb-2005.) |
Ref | Expression |
---|---|
elif | ⊢ (A ∈ if(φ, B, C) ↔ ((φ ∧ A ∈ B) ∨ (¬ φ ∧ A ∈ C))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eleq2 2414 | . 2 ⊢ ( if(φ, B, C) = B → (A ∈ if(φ, B, C) ↔ A ∈ B)) | |
2 | eleq2 2414 | . 2 ⊢ ( if(φ, B, C) = C → (A ∈ if(φ, B, C) ↔ A ∈ C)) | |
3 | 1, 2 | elimif 3692 | 1 ⊢ (A ∈ if(φ, B, C) ↔ ((φ ∧ A ∈ B) ∨ (¬ φ ∧ A ∈ C))) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 ↔ wb 176 ∨ wo 357 ∧ wa 358 ∈ wcel 1710 ifcif 3663 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 |
This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-clab 2340 df-cleq 2346 df-clel 2349 df-if 3664 |
This theorem is referenced by: enprmaplem4 6080 |
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