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Mirrors > Home > ILE Home > Th. List > eqsbc2 | GIF version |
Description: Substitution for the right-hand side in an equality. (Contributed by Alan Sare, 24-Oct-2011.) (Proof shortened by JJ, 7-Jul-2021.) |
Ref | Expression |
---|---|
eqsbc2 | ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqsbc1 3002 | . 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵)) | |
2 | eqcom 2179 | . . 3 ⊢ (𝐵 = 𝑥 ↔ 𝑥 = 𝐵) | |
3 | 2 | sbcbii 3022 | . 2 ⊢ ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐵) |
4 | eqcom 2179 | . 2 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵) | |
5 | 1, 3, 4 | 3bitr4g 223 | 1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 105 = wceq 1353 ∈ wcel 2148 [wsbc 2962 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-io 709 ax-5 1447 ax-7 1448 ax-gen 1449 ax-ie1 1493 ax-ie2 1494 ax-8 1504 ax-10 1505 ax-11 1506 ax-i12 1507 ax-bndl 1509 ax-4 1510 ax-17 1526 ax-i9 1530 ax-ial 1534 ax-i5r 1535 ax-ext 2159 |
This theorem depends on definitions: df-bi 117 df-tru 1356 df-nf 1461 df-sb 1763 df-clab 2164 df-cleq 2170 df-clel 2173 df-nfc 2308 df-v 2739 df-sbc 2963 |
This theorem is referenced by: (None) |
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