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Joined 2026-09-21 Message
Incoming links: Classical first-order logic
Classical first-order logic 2026-10-05
Classical first-order logic has the quantifier and logical equality rules of natural deduction together with unrestricted double-negation elimination, or equivalently the law of excluded middle. It is interpreted in ordinary two-valued first-order structures.
Intuitionistic first-order logic 2026-10-05
Intuitionistic first-order logic has the usual introduction and elimination rules for logical connectives, quantifiers and logical equality, but no unrestricted double-negation elimination or law of excluded middle. Its proofs retain the witness and logical disjunction information that can be lost in classical first-order logic.
Assume the existential statement and eliminate it using a fresh witness . Let abbreviate . Instantiating its universal quantification at the witness itself gives two incompatible conditional requirements.
1 | ∃x ∀y ((R(x,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y))) assume
2 | | ∀y ((R(u,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y))) fresh witness
3 | | (A → ¬A) ∧ (¬A → A) ∀-elim, 2, y := u
4 | | A → ¬A ∧-elim-L, 3
5 | | ¬A → A ∧-elim-R, 3
6 | | | A assume
7 | | | ¬A →-elim, 6 and 4
8 | | | ⊥ ¬-elim, 6 and 7
9 | | ¬A ¬-intro, discharge 6–8
10 | | A →-elim, 9 and 5
11 | | ⊥ ¬-elim, 10 and 9
12 | ⊥ ∃-elim, 1, subproof 2–11
13 ¬∃x ∀y ((R(x,y) → ¬R(y,y)) ∧ (¬R(y,y) → R(y,y))) ¬-intro, discharge 1–12The contradiction at line 12 contains no free witness variable, so the existential quantifier elimination rule meets its eigenvariable condition. This argument is already valid without classical first-order logic double-negation elimination.