Step
*
1
of Lemma
bpa-equiv-iff-norm
1. p : {2...}
2. EquivRel(basic-padic(p);x,y.bpa-equiv(p;x;y))
3. ∀x:basic-padic(p). bpa-equiv(p;x;bpa-norm(p;x))
4. x : basic-padic(p)
5. y : basic-padic(p)
6. bpa-equiv(p;x;y)
⊢ bpa-norm(p;x) = bpa-norm(p;y) ∈ basic-padic(p)
BY
{ ((D -3 THEN RenameVar `n' (-4) THEN RenameVar `a' (-3))
THEN D -2
THEN RenameVar `m' (-3)
THEN RenameVar `b' (-2)
THEN RepUR ``bpa-equiv`` -1
THEN RepUR ``bpa-norm`` 0
THEN AutoSplit) }
1
1. p : {2...}
2. EquivRel(basic-padic(p);x,y.bpa-equiv(p;x;y))
3. ∀x:basic-padic(p). bpa-equiv(p;x;bpa-norm(p;x))
4. n : ℕ
5. a : p-adics(p)
6. m : ℕ
7. b : p-adics(p)
8. p^m(p) * a = p^n(p) * b ∈ p-adics(p)
9. n = 0 ∈ ℤ
⊢ <0, a>
= if (m =z 0) then <0, b>
if (b m =z 0) then <0, p-shift(p;b;m)>
else let k,b = p-unitize(p;b;m)
in <m - k, b>
fi
∈ basic-padic(p)
2
1. p : {2...}
2. EquivRel(basic-padic(p);x,y.bpa-equiv(p;x;y))
3. ∀x:basic-padic(p). bpa-equiv(p;x;bpa-norm(p;x))
4. n : {1...}
5. a : p-adics(p)
6. m : ℕ
7. b : p-adics(p)
8. p^m(p) * a = p^n(p) * b ∈ p-adics(p)
⊢ if (a n =z 0) then <0, p-shift(p;a;n)> else let k,b = p-unitize(p;a;n) in <n - k, b> fi
= if (m =z 0) then <0, b>
if (b m =z 0) then <0, p-shift(p;b;m)>
else let k,b = p-unitize(p;b;m)
in <m - k, b>
fi
∈ basic-padic(p)
Latex:
Latex:
1. p : \{2...\}
2. EquivRel(basic-padic(p);x,y.bpa-equiv(p;x;y))
3. \mforall{}x:basic-padic(p). bpa-equiv(p;x;bpa-norm(p;x))
4. x : basic-padic(p)
5. y : basic-padic(p)
6. bpa-equiv(p;x;y)
\mvdash{} bpa-norm(p;x) = bpa-norm(p;y)
By
Latex:
((D -3 THEN RenameVar `n' (-4) THEN RenameVar `a' (-3))
THEN D -2
THEN RenameVar `m' (-3)
THEN RenameVar `b' (-2)
THEN RepUR ``bpa-equiv`` -1
THEN RepUR ``bpa-norm`` 0
THEN AutoSplit)
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