Step
*
2
of Lemma
do-apply-p-first
1. A : Type
2. B : Type
3. u : A ⟶ (B + Top)
4. v : (A ⟶ (B + Top)) List
5. ∀[x:A]. do-apply(p-first(v);x) = do-apply(hd(filter(λf.can-apply(f;x);v));x) ∈ B supposing ↑can-apply(p-first(v);x)
⊢ ∀[x:A]
do-apply(p-first([u / v]);x) = do-apply(hd(filter(λf.can-apply(f;x);[u / v]));x) ∈ B
supposing ↑can-apply(p-first([u / v]);x)
BY
{ Subst ⌜[u / v] ~ [u] @ v⌝ 0⋅ }
1
.....equality.....
1. A : Type
2. B : Type
3. u : A ⟶ (B + Top)
4. v : (A ⟶ (B + Top)) List
5. ∀[x:A]. do-apply(p-first(v);x) = do-apply(hd(filter(λf.can-apply(f;x);v));x) ∈ B supposing ↑can-apply(p-first(v);x)
⊢ [u / v] ~ [u] @ v
2
1. A : Type
2. B : Type
3. u : A ⟶ (B + Top)
4. v : (A ⟶ (B + Top)) List
5. ∀[x:A]. do-apply(p-first(v);x) = do-apply(hd(filter(λf.can-apply(f;x);v));x) ∈ B supposing ↑can-apply(p-first(v);x)
⊢ ∀[x:A]
do-apply(p-first([u] @ v);x) = do-apply(hd(filter(λf.can-apply(f;x);[u] @ v));x) ∈ B
supposing ↑can-apply(p-first([u] @ v);x)
Latex:
Latex:
1. A : Type
2. B : Type
3. u : A {}\mrightarrow{} (B + Top)
4. v : (A {}\mrightarrow{} (B + Top)) List
5. \mforall{}[x:A]
do-apply(p-first(v);x) = do-apply(hd(filter(\mlambda{}f.can-apply(f;x);v));x)
supposing \muparrow{}can-apply(p-first(v);x)
\mvdash{} \mforall{}[x:A]
do-apply(p-first([u / v]);x) = do-apply(hd(filter(\mlambda{}f.can-apply(f;x);[u / v]));x)
supposing \muparrow{}can-apply(p-first([u / v]);x)
By
Latex:
Subst \mkleeneopen{}[u / v] \msim{} [u] @ v\mkleeneclose{} 0\mcdot{}
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