Analyze failed or stuck proofs and propose auxiliary lemmas to help complete the proof in Isabelle/HOL or Coq. Use when encountering proof failures, stuck proof states, unprovable subgoals, or when needing to strengthen induction hypotheses. Identifies missing lemmas, suggests proof strategies, and generates helper lemmas with appropriate statements and proof sketches. Supports inductive proofs, case analysis, rewriting, and complex proof obligations.
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npx tessl i github:ArabelaTso/Skills-4-SE --skill lemma-discovery-assistant94
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Analyze stuck or failed proofs and propose auxiliary lemmas that can help complete the proof.
When proofs fail or get stuck, the issue is often a missing auxiliary lemma. This skill helps identify what lemmas are needed by:
Unprovable subgoal:
Weak induction hypothesis:
Missing intermediate steps:
Insufficient rewrite rules:
Examine the current proof context:
What to look for:
Questions to ask:
Determine what's preventing progress:
Gap types:
Formulate precise lemma statement:
Lemma characteristics:
Example patterns:
(* Isabelle *)
lemma helper_name: "⟦ assumptions ⟧ ⟹ conclusion"
(* Coq *)
Lemma helper_name : assumptions → conclusion.Provide guidance on proving the lemma:
Strategy types:
Show how the lemma helps the main proof:
Problem: Induction hypothesis too weak.
Solution: Strengthen the induction hypothesis.
Example (Isabelle):
Stuck proof:
lemma "length (reverse xs) = length xs"
proof (induction xs)
case Nil
then show ?case by simp
next
case (Cons x xs)
(* Stuck: reverse (x # xs) = reverse xs @ [x]
but we don't have a lemma about length of append *)
then show ?case sorry
qedProposed lemma:
lemma length_append: "length (xs @ ys) = length xs + length ys"
proof (induction xs)
case Nil
then show ?case by simp
next
case (Cons x xs)
then show ?case by simp
qedUsage: Apply length_append in the induction step.
Problem: Statement too specific to prove by induction.
Solution: Generalize with accumulator or additional parameter.
Example (Coq):
Stuck proof:
Fixpoint reverse {A} (l : list A) : list A :=
match l with
| [] => []
| x :: xs => reverse xs ++ [x]
end.
Lemma reverse_involutive : forall A (l : list A),
reverse (reverse l) = l.
Proof.
induction l.
- reflexivity.
- simpl. (* Stuck: need reverse (reverse l ++ [a]) = a :: l *)
Abort.Proposed lemma:
Lemma reverse_append : forall A (l1 l2 : list A),
reverse (l1 ++ l2) = reverse l2 ++ reverse l1.
Proof.
induction l1; intros; simpl.
- rewrite app_nil_r. reflexivity.
- rewrite IHl1. rewrite app_assoc. reflexivity.
Qed.Usage: Use reverse_append to handle the append in the goal.
Problem: Simplification doesn't reach desired form.
Solution: Add rewrite rules for specific patterns.
Example (Isabelle):
Stuck proof:
lemma "map f (map g xs) = map (f ∘ g) xs"
proof (induction xs)
case Nil
then show ?case by simp
next
case (Cons x xs)
(* Need: (f ∘ g) x = f (g x) *)
then show ?case sorry
qedProposed lemma:
lemma comp_apply: "(f ∘ g) x = f (g x)"
by (simp add: comp_def)Usage: Add to simplification set or apply explicitly.
Problem: Need properties about data structure operations.
Solution: Prove fundamental properties of the structure.
Example (Coq):
Stuck proof:
Fixpoint insert (x : nat) (t : tree) : tree := (* ... *)
Lemma insert_member : forall x t,
member x (insert x t) = true.
Proof.
induction t.
- (* Stuck: need properties of member and insert *)
Abort.Proposed lemmas:
Lemma member_insert_eq : forall x y t,
x = y → member x (insert y t) = true.
Lemma member_insert_neq : forall x y t,
x ≠ y → member x (insert y t) = member x t.
Lemma member_insert : forall x y t,
member x (insert y t) = (x =? y) || member x t.Usage: Case split on x = y and apply appropriate lemma.
For Isabelle-specific lemma patterns, tactics, and the sledgehammer tool, see references/isabelle_lemmas.md.
Key features:
try and try0 for tactic explorationFor Coq-specific lemma patterns, tactics, and proof search, see references/coq_lemmas.md.
Key features:
Search and SearchAbout for finding lemmasauto, eauto for proof searchLtacFor detailed lemma patterns organized by category, see references/proof_patterns.md.
Categories include:
Scenario: Proving correctness of insertion sort.
Main theorem (Isabelle):
lemma insertion_sort_correct:
"sorted (insertion_sort xs) ∧ mset (insertion_sort xs) = mset xs"Stuck proof:
proof (induction xs)
case Nil
then show ?case by simp
next
case (Cons x xs)
(* Need to show:
sorted (insert x (insertion_sort xs))
mset (insert x (insertion_sort xs)) = mset (x # xs) *)
then show ?case sorry
qedAnalysis:
insert functioninsert preserves sortinginsert preserves multisetProposed lemmas:
(* Lemma 1: insert preserves sorting *)
lemma insert_sorted:
"sorted xs ⟹ sorted (insert x xs)"
proof (induction xs)
case Nil
then show ?case by simp
next
case (Cons y ys)
then show ?case
by (auto simp: sorted_Cons)
qed
(* Lemma 2: insert preserves multiset *)
lemma mset_insert:
"mset (insert x xs) = {#x#} + mset xs"
proof (induction xs)
case Nil
then show ?case by simp
next
case (Cons y ys)
then show ?case by (auto simp: ac_simps)
qed
(* Lemma 3: sorted definition for cons *)
lemma sorted_Cons:
"sorted (x # xs) = (sorted xs ∧ (∀y ∈ set xs. x ≤ y))"
by (auto simp: sorted_append)Revised main proof:
lemma insertion_sort_correct:
"sorted (insertion_sort xs) ∧ mset (insertion_sort xs) = mset xs"
proof (induction xs)
case Nil
then show ?case by simp
next
case (Cons x xs)
have "sorted (insertion_sort xs)" using Cons by simp
then have "sorted (insert x (insertion_sort xs))"
by (rule insert_sorted)
moreover have "mset (insert x (insertion_sort xs)) = {#x#} + mset (insertion_sort xs)"
by (rule mset_insert)
moreover have "mset (insertion_sort xs) = mset xs"
using Cons by simp
ultimately show ?case by (auto simp: ac_simps)
qedPossible issues:
Solutions:
Possible issues:
Solutions:
For detailed guidance on specific aspects:
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