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Mirrors > Home > ILE Home > Th. List > eqsbc2 | Unicode 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 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqsbc1 2989 | . 2 | |
2 | eqcom 2167 | . . 3 | |
3 | 2 | sbcbii 3009 | . 2 |
4 | eqcom 2167 | . 2 | |
5 | 1, 3, 4 | 3bitr4g 222 | 1 |
Colors of variables: wff set class |
Syntax hints: wi 4 wb 104 wceq 1343 wcel 2136 wsbc 2950 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 699 ax-5 1435 ax-7 1436 ax-gen 1437 ax-ie1 1481 ax-ie2 1482 ax-8 1492 ax-10 1493 ax-11 1494 ax-i12 1495 ax-bndl 1497 ax-4 1498 ax-17 1514 ax-i9 1518 ax-ial 1522 ax-i5r 1523 ax-ext 2147 |
This theorem depends on definitions: df-bi 116 df-tru 1346 df-nf 1449 df-sb 1751 df-clab 2152 df-cleq 2158 df-clel 2161 df-nfc 2296 df-v 2727 df-sbc 2951 |
This theorem is referenced by: (None) |
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