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Mirrors > Home > MPE Home > Th. List > Mathboxes > fnopabeqd | Structured version Visualization version GIF version |
Description: Equality deduction for function abstractions. (Contributed by Jeff Madsen, 19-Jun-2011.) |
Ref | Expression |
---|---|
fnopabeqd.1 | ⊢ (𝜑 → 𝐵 = 𝐶) |
Ref | Expression |
---|---|
fnopabeqd | ⊢ (𝜑 → {〈𝑥, 𝑦〉 ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵)} = {〈𝑥, 𝑦〉 ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶)}) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | fnopabeqd.1 | . . . 4 ⊢ (𝜑 → 𝐵 = 𝐶) | |
2 | 1 | eqeq2d 2620 | . . 3 ⊢ (𝜑 → (𝑦 = 𝐵 ↔ 𝑦 = 𝐶)) |
3 | 2 | anbi2d 736 | . 2 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶))) |
4 | 3 | opabbidv 4648 | 1 ⊢ (𝜑 → {〈𝑥, 𝑦〉 ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵)} = {〈𝑥, 𝑦〉 ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶)}) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 ∈ wcel 1977 {copab 4642 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-opab 4644 |
This theorem is referenced by: (None) |
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