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Theorem ifandc 3541
 Description: Rewrite a conjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifandc DECID

Proof of Theorem ifandc
StepHypRef Expression
1 df-dc 821 . 2 DECID
2 iftrue 3510 . . . 4
3 ibar 299 . . . . 5
43ifbid 3526 . . . 4
52, 4eqtr2d 2191 . . 3
6 simpl 108 . . . . . 6
76con3i 622 . . . . 5
87iffalsed 3515 . . . 4
9 iffalse 3513 . . . 4
108, 9eqtr4d 2193 . . 3
115, 10jaoi 706 . 2
121, 11sylbi 120 1 DECID
 Colors of variables: wff set class Syntax hints:   wn 3   wi 4   wa 103   wo 698  DECID wdc 820   wceq 1335  cif 3505 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-in1 604  ax-in2 605  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-11 1486  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139 This theorem depends on definitions:  df-bi 116  df-dc 821  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-if 3506 This theorem is referenced by:  isumss  11270
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