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Theorem ax11v2 1992
 Description: Recovery of ax-11o 2141 from ax11v 2096. This proof uses ax-10 2140 and ax-11 1746. TODO: figure out if this is useful, or if it should be simplified or eliminated. (Contributed by NM, 2-Feb-2007.)
Hypothesis
Ref Expression
ax11v2.1 (x = z → (φx(x = zφ)))
Assertion
Ref Expression
ax11v2 x x = y → (x = y → (φx(x = yφ))))
Distinct variable groups:   x,z   y,z   φ,z
Allowed substitution hints:   φ(x,y)

Proof of Theorem ax11v2
StepHypRef Expression
1 a9ev 1656 . 2 z z = y
2 ax11v2.1 . . . . 5 (x = z → (φx(x = zφ)))
3 equequ2 1686 . . . . . . 7 (z = y → (x = zx = y))
43adantl 452 . . . . . 6 ((¬ x x = y z = y) → (x = zx = y))
5 dveeq2 1940 . . . . . . . . 9 x x = y → (z = yx z = y))
65imp 418 . . . . . . . 8 ((¬ x x = y z = y) → x z = y)
7 nfa1 1788 . . . . . . . . 9 xx z = y
83imbi1d 308 . . . . . . . . . 10 (z = y → ((x = zφ) ↔ (x = yφ)))
98sps 1754 . . . . . . . . 9 (x z = y → ((x = zφ) ↔ (x = yφ)))
107, 9albid 1772 . . . . . . . 8 (x z = y → (x(x = zφ) ↔ x(x = yφ)))
116, 10syl 15 . . . . . . 7 ((¬ x x = y z = y) → (x(x = zφ) ↔ x(x = yφ)))
1211imbi2d 307 . . . . . 6 ((¬ x x = y z = y) → ((φx(x = zφ)) ↔ (φx(x = yφ))))
134, 12imbi12d 311 . . . . 5 ((¬ x x = y z = y) → ((x = z → (φx(x = zφ))) ↔ (x = y → (φx(x = yφ)))))
142, 13mpbii 202 . . . 4 ((¬ x x = y z = y) → (x = y → (φx(x = yφ))))
1514ex 423 . . 3 x x = y → (z = y → (x = y → (φx(x = yφ)))))
1615exlimdv 1636 . 2 x x = y → (z z = y → (x = y → (φx(x = yφ)))))
171, 16mpi 16 1 x x = y → (x = y → (φx(x = yφ))))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃wex 1541 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925 This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545 This theorem is referenced by:  ax11a2  1993
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