DEFINITION r_dis()
TYPE =
∀k:K
.∀P:Prop
.(∀i:nat.(eq nat (r k i) i))→P
→((∀i:nat.(eq nat (r k i) (S i)))→P)→P
BODY =
assume k: K
we proceed by induction on k to prove
∀P:Prop
.(∀i:nat.(eq nat (r k i) i))→P
→((∀i:nat.(eq nat (r k i) (S i)))→P)→P
case Bind : b:B ⇒
the thesis becomes
∀P:Prop
.∀H:(∀i:nat.(eq nat (r (Bind b) i) i))→P
.((∀i:nat.(eq nat (r (Bind b) i) (S i)))→P)→P
assume P: Prop
suppose H: (∀i:nat.(eq nat (r (Bind b) i) i))→P
suppose : (∀i:nat.(eq nat (r (Bind b) i) (S i)))→P
assume i: nat
by (refl_equal . .)
we proved eq nat i i
that is equivalent to eq nat (r (Bind b) i) i
we proved ∀i:nat.(eq nat (r (Bind b) i) i)
by (H previous)
we proved P
∀P:Prop
.∀H:(∀i:nat.(eq nat (r (Bind b) i) i))→P
.((∀i:nat.(eq nat (r (Bind b) i) (S i)))→P)→P
case Flat : f:F ⇒
the thesis becomes
∀P:Prop
.(∀i:nat.(eq nat (r (Flat f) i) i))→P
→∀H0:(∀i:nat.(eq nat (r (Flat f) i) (S i)))→P.P
assume P: Prop
suppose : (∀i:nat.(eq nat (r (Flat f) i) i))→P
suppose H0: (∀i:nat.(eq nat (r (Flat f) i) (S i)))→P
assume i: nat
by (refl_equal . .)
we proved eq nat (S i) (S i)
that is equivalent to eq nat (r (Flat f) i) (S i)
we proved ∀i:nat.(eq nat (r (Flat f) i) (S i))
by (H0 previous)
we proved P
∀P:Prop
.(∀i:nat.(eq nat (r (Flat f) i) i))→P
→∀H0:(∀i:nat.(eq nat (r (Flat f) i) (S i)))→P.P
we proved
∀P:Prop
.(∀i:nat.(eq nat (r k i) i))→P
→((∀i:nat.(eq nat (r k i) (S i)))→P)→P
we proved
∀k:K
.∀P:Prop
.(∀i:nat.(eq nat (r k i) i))→P
→((∀i:nat.(eq nat (r k i) (S i)))→P)→P