Step
*
1
2
2
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. 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)
9. ¬(m = 0 ∈ ℤ)
⊢ 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)
BY
{ (OReduce 0 THENA Auto) }
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 : {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)
9. ¬(m = 0 ∈ ℤ)
⊢ 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 (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. n : \{1...\}
5. a : p-adics(p)
6. m : \mBbbN{}
7. b : p-adics(p)
8. p\^{}m(p) * a = p\^{}n(p) * b
9. \mneg{}(m = 0)
\mvdash{} if (a n =\msubz{} 0) then ɘ, p-shift(p;a;n)> else let k,b = p-unitize(p;a;n) in <n - k, b> fi
= if (m =\msubz{} 0) then ɘ, b>
if (b m =\msubz{} 0) then ɘ, p-shift(p;b;m)>
else let k,b = p-unitize(p;b;m)
in <m - k, b>
fi
By
Latex:
(OReduce 0 THENA Auto)
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