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Can one create such function in Agda ?

ℕ→ℕ-undecidable : ¬ ( (f g :  ℕ → ℕ ) → Dec (f ≡ g)) 
ℕ→ℕ-undecidable = ?

I am particularly interested in proof using cubical Agda.

András Kovács
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MJG
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2 Answers2

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ℕ→ℕ-undecidable is not provable in Agda. If we postulate the law of excluded middle (LEM), it follows that equality on every set is decidable, contradicting ℕ→ℕ-undecidable. Since Agda is consistent with LEM, it follows that ℕ→ℕ-undecidable is not provable in base Agda. This holds the same for cubical and vanilla Agda.

András Kovács
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The flag --injective-type-constructors makes Agda anti-classical. The proof of this can be seen here. See also this question on the Proof Assitants StackExchange.

Hjulle
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