Step
*
of Lemma
num-antecedents-fun_exp
∀[Info:Type]. ∀[es:EO+(Info)]. ∀[Sys:EClass(Top)]. ∀[f:sys-antecedent(es;Sys)]. ∀[n:ℕ]. ∀[e:E(Sys)].
#f(f^n e) = (#f(e) - n) ∈ ℤ supposing n ≤ #f(e)
BY
{ ((InductionOnNat⋅ THEN Reduce 0 THEN Auto)
THEN (Assert #f(e) = (1 + #f(f e)) ∈ ℤ BY
(MoveToConcl (-1)
THEN RW (AddrC [2;2] (RecUnfoldC `num-antecedents`)) 0
THEN RW (AddrC [1;2] (RecUnfoldC `num-antecedents`)) 0
THEN AutoSplit))
) }
1
1. Info : Type
2. es : EO+(Info)
3. Sys : EClass(Top)
4. f : sys-antecedent(es;Sys)
5. n : ℤ
6. 0 < n
7. ∀[e:E(Sys)]. #f(f^n - 1 e) = (#f(e) - n - 1) ∈ ℤ supposing (n - 1) ≤ #f(e)
8. e : E(Sys)
9. n ≤ #f(e)
10. #f(e) = (1 + #f(f e)) ∈ ℤ
⊢ #f(f^n e) = (#f(e) - n) ∈ ℤ
Latex:
\mforall{}[Info:Type]. \mforall{}[es:EO+(Info)]. \mforall{}[Sys:EClass(Top)]. \mforall{}[f:sys-antecedent(es;Sys)]. \mforall{}[n:\mBbbN{}]. \mforall{}[e:E(Sys)].
\#f(f\^{}n e) = (\#f(e) - n) supposing n \mleq{} \#f(e)
By
((InductionOnNat\mcdot{} THEN Reduce 0 THEN Auto)
THEN (Assert \#f(e) = (1 + \#f(f e)) BY
(MoveToConcl (-1)
THEN RW (AddrC [2;2] (RecUnfoldC `num-antecedents`)) 0
THEN RW (AddrC [1;2] (RecUnfoldC `num-antecedents`)) 0
THEN AutoSplit))
)
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