lean2/tests/lean/tst6.lean.expected.out
Leonardo de Moura bff5a6bfb2 fix(frontends/lean/pp): make sure pp and parser are using the same precedences
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
2013-12-19 12:46:14 -08:00

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Set: pp::colors
Set: pp::unicode
Assumed: N
Assumed: h
Proved: CongrH
Set: lean::pp::implicit
Theorem CongrH {a1 a2 b1 b2 : N} (H1 : eq::explicit N a1 b1) (H2 : eq::explicit N a2 b2) : eq::explicit
N
(h a1 a2)
(h b1 b2) :=
Congr::explicit
N
(λ x : N, N)
(h a1)
(h b1)
a2
b2
(Congr::explicit N (λ x : N, N → N) h h a1 b1 (Refl::explicit (N → N → N) h) H1)
H2
Theorem CongrH::explicit (a1 a2 b1 b2 : N) (H1 : a1 = b1) (H2 : a2 = b2) : h a1 a2 = h b1 b2 := CongrH H1 H2
Set: lean::pp::implicit
Theorem CongrH {a1 a2 b1 b2 : N} (H1 : a1 = b1) (H2 : a2 = b2) : h a1 a2 = h b1 b2 := Congr (Congr (Refl h) H1) H2
Theorem CongrH::explicit (a1 a2 b1 b2 : N) (H1 : a1 = b1) (H2 : a2 = b2) : h a1 a2 = h b1 b2 := CongrH H1 H2
Proved: Example1
Set: lean::pp::implicit
Theorem Example1 (a b c d : N)
(H : eq::explicit N a b ∧ eq::explicit N b c eq::explicit N a d ∧ eq::explicit N d c) : eq::explicit
N
(h a b)
(h c b) :=
DisjCases::explicit
(eq::explicit N a b ∧ eq::explicit N b c)
(eq::explicit N a d ∧ eq::explicit N d c)
(h a b == h c b)
H
(λ H1 : eq::explicit N a b ∧ eq::explicit N b c,
CongrH::explicit
a
b
c
b
(Trans::explicit
N
a
b
c
(Conjunct1::explicit (eq::explicit N a b) (eq::explicit N b c) H1)
(Conjunct2::explicit (eq::explicit N a b) (eq::explicit N b c) H1))
(Refl::explicit N b))
(λ H1 : eq::explicit N a d ∧ eq::explicit N d c,
CongrH::explicit
a
b
c
b
(Trans::explicit
N
a
d
c
(Conjunct1::explicit (eq::explicit N a d) (eq::explicit N d c) H1)
(Conjunct2::explicit (eq::explicit N a d) (eq::explicit N d c) H1))
(Refl::explicit N b))
Proved: Example2
Set: lean::pp::implicit
Theorem Example2 (a b c d : N)
(H : eq::explicit N a b ∧ eq::explicit N b c eq::explicit N a d ∧ eq::explicit N d c) : eq::explicit
N
(h a b)
(h c b) :=
DisjCases::explicit
(eq::explicit N a b ∧ eq::explicit N b c)
(eq::explicit N a d ∧ eq::explicit N d c)
(eq::explicit N (h a b) (h c b))
H
(λ H1 : eq::explicit N a b ∧ eq::explicit N b c,
CongrH::explicit
a
b
c
b
(Trans::explicit
N
a
b
c
(Conjunct1::explicit (a == b) (b == c) H1)
(Conjunct2::explicit (a == b) (b == c) H1))
(Refl::explicit N b))
(λ H1 : eq::explicit N a d ∧ eq::explicit N d c,
CongrH::explicit
a
b
c
b
(Trans::explicit
N
a
d
c
(Conjunct1::explicit (a == d) (d == c) H1)
(Conjunct2::explicit (a == d) (d == c) H1))
(Refl::explicit N b))
Proved: Example3
Set: lean::pp::implicit
Theorem Example3 (a b c d e : N) (H : a = b ∧ b = e ∧ b = c a = d ∧ d = c) : h a b = h c b :=
DisjCases
H
(λ H1 : a = b ∧ b = e ∧ b = c, CongrH (Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1))) (Refl b))
(λ H1 : a = d ∧ d = c, CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
Proved: Example4
Set: lean::pp::implicit
Theorem Example4 (a b c d e : N) (H : a = b ∧ b = e ∧ b = c a = d ∧ d = c) : h a c = h c a :=
DisjCases
H
(λ H1 : a = b ∧ b = e ∧ b = c,
let AeqC := Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1)) in CongrH AeqC (Symm AeqC))
(λ H1 : a = d ∧ d = c, let AeqC := Trans (Conjunct1 H1) (Conjunct2 H1) in CongrH AeqC (Symm AeqC))