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Theorem reu7 2707
Description: Restricted uniqueness using implicit substitution. (Contributed by NM, 24-Oct-2006.)
Hypothesis
Ref Expression
rmo4.1 (x = y → (φψ))
Assertion
Ref Expression
reu7 (∃!x A φ ↔ (x A φ x A y A (ψx = y)))
Distinct variable groups:   x,y,A   φ,y   ψ,x
Allowed substitution hints:   φ(x)   ψ(y)

Proof of Theorem reu7
Dummy variable z is distinct from all other variables.
StepHypRef Expression
1 reu3 2702 . 2 (∃!x A φ ↔ (x A φ z A x A (φx = z)))
2 rmo4.1 . . . . . . 7 (x = y → (φψ))
3 equequ1 1574 . . . . . . . 8 (x = y → (x = zy = z))
4 equcom 1569 . . . . . . . 8 (y = zz = y)
53, 4syl6bb 185 . . . . . . 7 (x = y → (x = zz = y))
62, 5imbi12d 223 . . . . . 6 (x = y → ((φx = z) ↔ (ψz = y)))
76cbvralv 2505 . . . . 5 (x A (φx = z) ↔ y A (ψz = y))
87rexbii 2303 . . . 4 (z A x A (φx = z) ↔ z A y A (ψz = y))
9 equequ1 1574 . . . . . . 7 (z = x → (z = yx = y))
109imbi2d 219 . . . . . 6 (z = x → ((ψz = y) ↔ (ψx = y)))
1110ralbidv 2298 . . . . 5 (z = x → (y A (ψz = y) ↔ y A (ψx = y)))
1211cbvrexv 2506 . . . 4 (z A y A (ψz = y) ↔ x A y A (ψx = y))
138, 12bitri 173 . . 3 (z A x A (φx = z) ↔ x A y A (ψx = y))
1413anbi2i 430 . 2 ((x A φ z A x A (φx = z)) ↔ (x A φ x A y A (ψx = y)))
151, 14bitri 173 1 (∃!x A φ ↔ (x A φ x A y A (ψx = y)))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98  wral 2278  wrex 2279  ∃!wreu 2280
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-io 614  ax-5 1312  ax-7 1313  ax-gen 1314  ax-ie1 1358  ax-ie2 1359  ax-8 1371  ax-10 1372  ax-11 1373  ax-i12 1374  ax-bnd 1375  ax-4 1376  ax-17 1395  ax-i9 1399  ax-ial 1403  ax-i5r 1404  ax-ext 1998
This theorem depends on definitions:  df-bi 110  df-tru 1229  df-nf 1326  df-sb 1622  df-eu 1879  df-mo 1880  df-cleq 2009  df-clel 2012  df-nfc 2143  df-ral 2283  df-rex 2284  df-reu 2285  df-rmo 2286
This theorem is referenced by: (None)
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