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Theorem clf 105
Description: Evaluate a lambda expression.
Hypotheses
Ref Expression
clf.1 A:β
clf.2 C:α
clf.3 [x:α = C]⊧[A = B]
clf.4 ⊤⊧[(λx:α By:α) = B]
clf.5 ⊤⊧[(λx:α Cy:α) = C]
Assertion
Ref Expression
clf ⊤⊧[(λx:α AC) = B]
Distinct variable groups:   y,A   y,B   y,C   x,y,α

Proof of Theorem clf
StepHypRef Expression
1 clf.2 . 2 C:α
2 clf.1 . . . . 5 A:β
32wl 59 . . . 4 λx:α A:(αβ)
43, 1wc 45 . . 3 (λx:α AC):β
5 clf.3 . . . 4 [x:α = C]⊧[A = B]
62, 5eqtypi 69 . . 3 B:β
74, 6weqi 68 . 2 [(λx:α AC) = B]:∗
8 clf.4 . . . 4 ⊤⊧[(λx:α By:α) = B]
98ax-cb1 29 . . 3 ⊤:∗
102beta 82 . . 3 ⊤⊧[(λx:α Ax:α) = A]
119, 10a1i 28 . 2 ⊤⊧[(λx:α Ax:α) = A]
12 weq 38 . . 3 = :(β → (β → ∗))
13 wv 58 . . 3 y:α:α
1412, 13, 9a17i 96 . . 3 ⊤⊧[(λx:α = y:α) = = ]
152, 13, 9hbl1 94 . . . 4 ⊤⊧[(λx:α λx:α Ay:α) = λx:α A]
16 clf.5 . . . 4 ⊤⊧[(λx:α Cy:α) = C]
173, 1, 13, 15, 16hbc 100 . . 3 ⊤⊧[(λx:α (λx:α AC)y:α) = (λx:α AC)]
1812, 4, 13, 6, 14, 17, 8hbov 101 . 2 ⊤⊧[(λx:α [(λx:α AC) = B]y:α) = [(λx:α AC) = B]]
19 wv 58 . . . 4 x:α:α
203, 19wc 45 . . 3 (λx:α Ax:α):β
2119, 1weqi 68 . . . . 5 [x:α = C]:∗
2221id 25 . . . 4 [x:α = C]⊧[x:α = C]
233, 19, 22ceq2 80 . . 3 [x:α = C]⊧[(λx:α Ax:α) = (λx:α AC)]
2412, 20, 2, 23, 5oveq12 90 . 2 [x:α = C]⊧[[(λx:α Ax:α) = A] = [(λx:α AC) = B]]
251, 7, 11, 18, 24insti 104 1 ⊤⊧[(λx:α AC) = B]
Colors of variables: type var term
Syntax hints:  tv 1  ht 2  hb 3  kc 5  λkl 6   = ke 7  kt 8  [kbr 9  wffMMJ2 11  wffMMJ2t 12
This theorem was proved from axioms:  ax-syl 15  ax-jca 17  ax-simpl 20  ax-simpr 21  ax-id 24  ax-trud 26  ax-cb1 29  ax-cb2 30  ax-refl 39  ax-eqmp 42  ax-ceq 46  ax-beta 60  ax-distrc 61  ax-leq 62  ax-hbl1 93  ax-17 95  ax-inst 103
This theorem depends on definitions:  df-ov 65
This theorem is referenced by:  cl  106  cbvf  167  exmid  186  axrep  207
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