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Mirrors > Home > ILE Home > Th. List > 2sb5 | GIF version |
Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.) |
Ref | Expression |
---|---|
2sb5 | ⊢ ([z / x][w / y]φ ↔ ∃x∃y((x = z ∧ y = w) ∧ φ)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sb5 1764 | . 2 ⊢ ([z / x][w / y]φ ↔ ∃x(x = z ∧ [w / y]φ)) | |
2 | 19.42v 1783 | . . . 4 ⊢ (∃y(x = z ∧ (y = w ∧ φ)) ↔ (x = z ∧ ∃y(y = w ∧ φ))) | |
3 | anass 381 | . . . . 5 ⊢ (((x = z ∧ y = w) ∧ φ) ↔ (x = z ∧ (y = w ∧ φ))) | |
4 | 3 | exbii 1493 | . . . 4 ⊢ (∃y((x = z ∧ y = w) ∧ φ) ↔ ∃y(x = z ∧ (y = w ∧ φ))) |
5 | sb5 1764 | . . . . 5 ⊢ ([w / y]φ ↔ ∃y(y = w ∧ φ)) | |
6 | 5 | anbi2i 430 | . . . 4 ⊢ ((x = z ∧ [w / y]φ) ↔ (x = z ∧ ∃y(y = w ∧ φ))) |
7 | 2, 4, 6 | 3bitr4ri 202 | . . 3 ⊢ ((x = z ∧ [w / y]φ) ↔ ∃y((x = z ∧ y = w) ∧ φ)) |
8 | 7 | exbii 1493 | . 2 ⊢ (∃x(x = z ∧ [w / y]φ) ↔ ∃x∃y((x = z ∧ y = w) ∧ φ)) |
9 | 1, 8 | bitri 173 | 1 ⊢ ([z / x][w / y]φ ↔ ∃x∃y((x = z ∧ y = w) ∧ φ)) |
Colors of variables: wff set class |
Syntax hints: ∧ wa 97 ↔ wb 98 ∃wex 1378 [wsb 1642 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1333 ax-gen 1335 ax-ie1 1379 ax-ie2 1380 ax-8 1392 ax-11 1394 ax-4 1397 ax-17 1416 ax-i9 1420 ax-ial 1424 |
This theorem depends on definitions: df-bi 110 df-sb 1643 |
This theorem is referenced by: opelopabsbALT 3987 |
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