Step
*
1
1
of Lemma
proof-tree-induction
.....antecedent.....
1. [Sequent] : Type
2. [Rule] : Type
3. effect : (Sequent × Rule) ⟶ (Sequent List?)
4. [Q] : proof-tree(Sequent;Rule;effect) ⟶ ℙ
5. ∀s:Sequent. ∀r:Rule. Q[proof-abort(s;r)] supposing ↑isr(effect <s, r>)
6. ∀s:Sequent. ∀r:Rule.
∀L:proof-tree(Sequent;Rule;effect) List
(∀pf∈L.Q[pf])
⇒ Q[make-proof-tree(s;r;L)] supposing ||L|| = ||outl(effect <s, r>)|| ∈ ℤ
supposing ↑isl(effect <s, r>)
7. pf : proof-tree(Sequent;Rule;effect)
8. s : Sequent
9. r : Rule
10. x : Sequent List
11. (effect <s, r>) = (inl x) ∈ (Sequent List?)
12. ∀L:proof-tree(Sequent;Rule;effect) List. (∀pf∈L.Q[pf])
⇒ Q[make-proof-tree(s;r;L)] supposing ||L|| = ||x|| ∈ ℤ
13. f : ℕ||x|| ⟶ proof-tree(Sequent;Rule;effect)
14. ∀b:ℕ||x||. Q[f b]
⊢ (∀pf∈mklist(||x||;f).Q[pf])
BY
{ (D 0 THENA Auto) }
1
1. [Sequent] : Type
2. [Rule] : Type
3. effect : (Sequent × Rule) ⟶ (Sequent List?)
4. [Q] : proof-tree(Sequent;Rule;effect) ⟶ ℙ
5. ∀s:Sequent. ∀r:Rule. Q[proof-abort(s;r)] supposing ↑isr(effect <s, r>)
6. ∀s:Sequent. ∀r:Rule.
∀L:proof-tree(Sequent;Rule;effect) List
(∀pf∈L.Q[pf])
⇒ Q[make-proof-tree(s;r;L)] supposing ||L|| = ||outl(effect <s, r>)|| ∈ ℤ
supposing ↑isl(effect <s, r>)
7. pf : proof-tree(Sequent;Rule;effect)
8. s : Sequent
9. r : Rule
10. x : Sequent List
11. (effect <s, r>) = (inl x) ∈ (Sequent List?)
12. ∀L:proof-tree(Sequent;Rule;effect) List. (∀pf∈L.Q[pf])
⇒ Q[make-proof-tree(s;r;L)] supposing ||L|| = ||x|| ∈ ℤ
13. f : ℕ||x|| ⟶ proof-tree(Sequent;Rule;effect)
14. ∀b:ℕ||x||. Q[f b]
15. i : ℕ||mklist(||x||;f)||
⊢ Q[mklist(||x||;f)[i]]
Latex:
Latex:
.....antecedent.....
1. [Sequent] : Type
2. [Rule] : Type
3. effect : (Sequent \mtimes{} Rule) {}\mrightarrow{} (Sequent List?)
4. [Q] : proof-tree(Sequent;Rule;effect) {}\mrightarrow{} \mBbbP{}
5. \mforall{}s:Sequent. \mforall{}r:Rule. Q[proof-abort(s;r)] supposing \muparrow{}isr(effect <s, r>)
6. \mforall{}s:Sequent. \mforall{}r:Rule.
\mforall{}L:proof-tree(Sequent;Rule;effect) List
(\mforall{}pf\mmember{}L.Q[pf]) {}\mRightarrow{} Q[make-proof-tree(s;r;L)] supposing ||L|| = ||outl(effect <s, r>)||
supposing \muparrow{}isl(effect <s, r>)
7. pf : proof-tree(Sequent;Rule;effect)
8. s : Sequent
9. r : Rule
10. x : Sequent List
11. (effect <s, r>) = (inl x)
12. \mforall{}L:proof-tree(Sequent;Rule;effect) List
(\mforall{}pf\mmember{}L.Q[pf]) {}\mRightarrow{} Q[make-proof-tree(s;r;L)] supposing ||L|| = ||x||
13. f : \mBbbN{}||x|| {}\mrightarrow{} proof-tree(Sequent;Rule;effect)
14. \mforall{}b:\mBbbN{}||x||. Q[f b]
\mvdash{} (\mforall{}pf\mmember{}mklist(||x||;f).Q[pf])
By
Latex:
(D 0 THENA Auto)
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