CIS6700-Spring2025/Homework 2.md

N.B. The definition of multistep reduction I use is composed of a reflexive case (t →* t) and a transitive case (t₁ → t₂ →* t₃). I need the following two lemmas throughout.

Lemma (transitivity): If t₁ →* t₂ →* t₃ then t₁ →* t₃. Proof: By induction on t₁ →* t₂. The reflexive case is trivial and the transitive case uses the induction hypothesis to combine the two multistep reductions.

Lemma (congruence): Multistep reduction is congruent in head positions, i.e. if t₁ →* t₁' then if t₁ then t₂ else t₃ →* if t₁' then t₂ else t₃, and similarly for succ, pred, and iszero. Proof: If the reduction is reflexive, then the proof is trivial. Otherwise, the reduction is of the form t₁ → t₁' →* t₁''. By E-If (likewise E-Succ, E-Pred, E-Iszero), we have if t₁ then t₂ else t₃ → if t₁' then t₂ else t₃. By the induction hypothesis, we have if t₁' then t₂ else t₃ →* if t₁'' then t₂ else t₃ (likewise forsucc, pred, iszero). We have our goal by combining the two by transitivity.

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Exercise 3.5.16: The two treatments of run-time errors agree if the following hold:

1. If t evaluates to a stuck state (using the rules without E-*-Wrong), then it evaluates to wrong (using the rules with E-*-Wrong). More precisely, if t →* t' ↛ and t ≠ v, then t →* wrong.
2. If t evaluates to wrong and does not contain wrong as a subterm, then t evaluates to a stuck state. This is the converse of the first statement, except we exclude cases where t was wrong to begin with.

Lemma 0: If t ↛ (using the rules without E-*-Wrong) and t ≠ v then t → wrong (using the rules with E-*-Wrong). Proof: By induction on the shape of t. The cases where t is true, false, 0, or succ nv are impossible.

• Case t = if t₁ then t₂ else t₃: The term does not step by E-IfTrue, E-IfFalse, nor E-If, so t₁ is not true nor false, and does not step. If t₁ is a value, then it is either a numeric value nv; otherwise, if it is not a value, then by the induction hypothesis, it steps to wrong. In both cases, t₁ is a badbool, so the whole expression steps to wrong by E-If-Wrong.
• Case t = succ t₁ where t₁ ≠ nv: The term does not step by E-Succ, so t₁ does not step. If t₁ is a value, then it must be true or false; otherwise, if it is not a value, then by the induction hypothesis, it steps to wrong. In both cases, t₁ is a badnat, so the whole expression steps to wrong by E-Succ-Wrong.
• Case t = pred t₁: The term does not step by E-PredZero, E-PredSucc, nor E-Pred, so t₁ ≠ nv and does not step. Then the argument for the previous case follows, using instead E-Pred-Wrong.
• Case t = iszero t₁: The argument for the previous case follows, noting that the term does not step by E-IszeroZero, E-IszeroSucc, nor E-Iszero, and using E-Iszero-Wrong to step.

Lemma 1: If t → t' (using the rules with E-*-Wrong) and t' contains a wrong subterm, then either t contains a wrong subterm, or t ↛ (using the rules without E-*-Wrong) and t ≠ v. Proof: By induction on t → t'.

• Cases E-If-True, E-If-False: The reduction is if true then t₁ else t₂ → t₁ or if false then t₁ else t₂ → t₂. If the RHS contains a wrong subterm, then evidently the LHS does as well.
• Cases E-If, E-Pred, E-Iszero: The three cases are similar; I use E-Iszero as representative. The reduction is iszero t₁ → iszero t₂, where t₁ → t₂, and t₂ must contain a wrong subterm. By the induction hypothesis, either t₁ contains a wrong subterm, in which case so does iszero t₁, or t₁ ↛ and t₁ ≠ v. In this latter case, we have iszero t₁ ↛, since none of E-Iszero-Zero, E-Iszero-Succ, and E-Iszero apply, and iszero t₁ is obviously not a value.
• Case E-Succ: The reduction is succ t₁ → succ t₂, where t₁ → t₂, and t₂ must contain a wrong subterm. By the induction hypothesis, either t₁ contains a wrong subterm, in which case so does succ t₁, or t₁ ↛ and t₁ ≠ v. In this latter case, we have succ t₁ ↛ since E-Succ does not apply, and we know that succ t₁ is not a value since t₁ is not a value.

Lemma 2: If t →* wrong (using the rules with E-*-Wrong), then either t contains a wrong subterm, or t →* t' ↛ (using the rules without E-*-Wrong) and t' ≠ v. Proof: By induction on t →* wrong. The reflexive case wrong →* wrong is trivial. In the transitive case t → t' →* wrong, by the induction hypothesis, either t' contains a wrong subterm, or t' →* t'' ↛ and t'' ≠ v. In the former case, we use Lemma 1 on the reduction t → t'. In the latter case, we have t → t' →* t'' ↛ and t'' ≠ v as required.

Proof of 3.5.16 (Condition 1): Suppose t →* t' ↛ and t' ≠ v. Then t' → wrong by Lemma 0, and t →* wrong by transitivity. Proof of 3.5.16 (Condition 2): Suppose t →* wrong and t does not contain a wrong subterm. Then by Lemma 2, it must be that t →* t' ↛ and t' ≠ v.

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Exercise 3.5.17: t →* v iff t ⇓ v.

Lemma 3: If t₁ → t₂ and t₂ ⇓ v, then t₁ ⇓ v. Proof: By induction on t₁ → t₂.

• Case E-IfTrue: The reduction is if true then t₄ else t₅ → t₂, with t₂ ⇓ v. By B-Value, true ⇓ true. Then by B-IfTrue, we construct if true then t₄ else t₅ ⇓ v.
• Case E-IfFalse: Similar as above, but using B-IfFalse.
• Case E-If: The reduction is if t₃ then t₄ else t₅ → if t₃' then t₄ else t₅, where t₃ → t₃', with if t₃' then t₄ else t₅ ⇓ v. By inversion on the latter, the only possible cases are B-IfTrue and B-IfFalse.
  • Subcase B-IfTrue: We have t₃' ⇓ true and t₄ ⇓ v. By the induction hypothesis, we have t₃ ⇓ true. Then by B-IfTrue, we construct if t₃ then t₄ else t₄ ⇓ true.
  • Subcase B-IfFalse: Similar as above, but the induction hypothesis gives t₃ ⇓ false, and we use B-IfFalse instead.
• Case E-Succ: The reduction is succ t₁ → succ t₂, where t₁ → t₂, with succ t₂ ⇓ v. By inversion on the latter, the only possible case is B-Succ, with t₂ ⇓ v. By the induction hypothesis, we have t₁ ⇓ v. Then by B-Succ, we construct succ t₁ ⇓ v.
• Case E-PredZero: The reduction is pred 0 → 0, with 0 ⇓ v. By inversion on the latter, the only possible case is B-Value, with v = 0. The goal pred 0 ⇓ 0 holds directly by B-PredZero.
• Case E-PredSucc: The reduction is pred (succ nv) → nv, with nv ⇓ v. By inversion on the latter, the only possible case is B-Value, with v = nv. The goal pred (succ nv) ⇓ nv holds directly by B-PredSucc, whose premise succ nv ⇓ succ nv holds by B-succ, whose premise nv ⇓ nv holds by B-Value.
• Case E-Pred: The reduction is pred t₁ → pred t₂, where t₁ → t₂, with pred t₂ ⇓ v. By inversion on the latter, the only possible cases are B-PredZero and B-PredSucc.
  • Subcase B-PredZero: We have pred t₂ ⇓ 0, where t₂ ⇓ 0. By the induction hypothesis, we have t₁ ⇓ 0. Then by B-PredZero, we construct pred t₁ ⇓ 0.
  • Subcase B-PredSucc: We have pred t₂ ⇓ nv, where t₂ ⇓ succ nv. By the induction hypothesis, we have t₁ ⇓ succ nv. Then by B-PredSucc, we construct pred t₁ ⇓ nv.
• Case E-IszeroZero: The reduction is iszero 0 → true, with true ⇓ v. By inversion on the latter, the only possible case is B-Value, with v = true. The goal iszero 0 ⇓ true holds directly by B-IszeroZero, whose premise 0 ⇓ 0 holds by B-Value.
• Case E-IszeroSucc: The reduction is iszero (succ nv) → false, with false ⇓ v. By inversion on the latter, the only possible case is B-Value, with v = false. The goal iszero (succ nv) ⇓ false holds directly by B-IszeroSucc, whose premise succ nv ⇓ succ nv holds by B-Succ, whose premise nv ⇓ nv holds by B-Value.
• Case E-Iszero: The reduction is iszero t₁ → iszero t₂, where t₁ → t₂, with iszero t₂ ⇓ v. By inversion on the latter, the only possible cases are B-IszeroZero and B-IszeroSucc.
  • Subcase B-IszeroZero: We have iszero t₂ ⇓ true, where t₂ ⇓ 0. By the induction hypothesis, we have t₁ ⇓ 0. Then by B-IszeroZero, we construct iszero t₁ ⇓ true.
  • Subcase B-IszeroSucc: We have iszero t₂ ⇓ false, where t₂ ⇓ succ nv. By the induction hypothesis, we have t₁ ⇓ succ nv. Then by B-IszeroSucc, we construct iszero t₁ ⇓ false.

Proof of 3.5.17 (LtR direction): By induction on t →* v. In the reflexive case, we have v → v, and v ⇓ v holds by B-Value. In the transitive case, we have t₁ → t₂ →* v. By the induction hypothesis, we have t₂ ⇓ v. Then by Lemma 3, we have t₁ ⇓ v.

Proof of 3.5.17 (RtL direction): By induction on t ⇓ v. I freely use the transitivity and congruence lemmata to combine multistep reductions.

• Case B-Value: By reflexivity.
• Case B-IfTrue: The evaluation is if t₁ then t₂ else t₃ ⇓ v, where t₁ ⇓ true and t₂ ⇓ v. By the induction hypothesis, we have t₁ →* true and t₂ →* v. Then we have if t₁ then t₂ else →* if true then t₂ else t₃ →* t₂ →* v, the middle step by E-IfTrue.
• Case B-IfFalse: Similar as above, but using E-IfFalse.
• Case B-Succ: The evaluation is succ t ⇓ succ v, where t ⇓ v. By the induction hypothesis, we have t →* v. Then we have succ t →* succ v by congruence.
• Case B-PredZero: The evaluation is pred t ⇓ 0, where t ⇓ 0. By the induction hypothesis, we have t →* 0. Then we have pred t →* pred 0 →* 0, the last step by E-PredZero.
• Case B-PredSucc: The evaluation is pred t ⇓ nv, where t ⇓ succ nv. By the induction hypothesis, we have t →* succ nv. Then we have pred t →* pred (succ nv) →* nv, the last step by E-PredSucc.
• Case B-IszeroZero: The evaluation is iszero t ⇓ true, where t ⇓ 0. By the induction hypothesis, we have t →* 0. Then we have iszero t →* iszero 0 →* true, the last step by E-IszeroZero.
• Case B-IszeroSucc: The evaluation is iszero t ⇓ false, where t ⇓ succ nv. By the induction hypothesis, we have t →* succ nv. Then we have iszero t →* iszero (succ nv) →* false, the last step by E-IszeroSucc.