# Logical connectives: testing ∨, →, ↔ and brackets on divisibility statements [← Guide index](https://ocme.evemisslab.com/en/derivations/) · [Original MKO](https://ocme.evemisslab.com/en/mko/mko-logical-connectives/) This is an owner-adopted natural-language teaching guide. Finite/arithmetic checks are not universal proofs; no Lean was run, and review scope is recorded per revision. Original source status labels are retained as historical data. Formulas retain exact TeX; no new semantic AST is invented. derivation-mko-logical-connectives-01 · production-20261019-adopted-96 ## Goal and scope For an integer n let P(n) be '4∣n' and Q(n) be '2∣n'. (1) Decide the truth of '4∣12 ∨ 6∣12' and explain that disjunction is inclusive; (2) prove ∀n∈ℤ (P(n)→Q(n)); (3) prove that ∀n∈ℤ (Q(n)→P(n)) is false, and hence that ∀n∈ℤ (P(n)↔Q(n)) is false; (4) decide the truth of P(3)→Q(3) and of '2+2=4 → 7 is odd', explaining that implication is not causation; (5) compare the truth of ¬Q(2)∧P(2) and ¬(Q(2)∧P(2)). Uses the MKO's definitions of the connectives, read as truth conditions in an introductory classical two-valued context, to decide the truth of compound propositions. Only concrete statements about divisibility and parity are used; intuitionistic, modal or relevance logics are not treated, and the canonical MKO and its English edition are not changed. ## Givens and symbols - divides_def - imp_intro - forall_intro - quantifier_negation - order_basics - numeric_evaluation - G1 — Context: introductory classical two-valued logic; every proposition is exactly one of true (T) or false (F) (the context of mko-proposition). The connectives are as defined in the MKO. - G2 — Divisibility: for integers d and n, d∣n means that n=dk for some integer k. For every integer n, P(n) is '4∣n' and Q(n) is '2∣n'. - G3 — Concrete statements: '6∣12'; '2+2=4'; '7 is odd', where odd means equal to 2m+1 for some integer m. - \neg,\ \land,\ \lor — negation, conjunction, (inclusive) disjunction - \to,\ \leftrightarrow — (material) implication and the biconditional - P(n),\ Q(n) — 4∣n and 2∣n - d\mid n — there is an integer k with n=dk - \mathrm{T},\ \mathrm{F} — true, false ## Rules and sources - divides_def · prerequisite — Divisibility: for integers a, b, a∣b means there is an integer k with b=ak. - imp_intro · prerequisite — →-introduction (discharging a hypothesis): if Q is derived inside a scope that temporarily assumes P, then P⇒Q follows, and P may no longer be used as known. - forall_intro · prerequisite — ∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain. - quantifier_negation · prerequisite — Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x). - order_basics · prerequisite — Order of the reals: exactly one of uv holds; u0 give uwvw. - numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable. - exists_intro · prerequisite — ∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x). - exists_elim · prerequisite — ∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a. - mul_assoc · prerequisite — Associativity of multiplication: (uv)w=u(vw). - case_analysis · prerequisite — Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility. - mul_zero · prerequisite — Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0. - mul_one · prerequisite — Multiplicative identity: 1·u=u. - contradiction · prerequisite — Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds. - iff_intro · prerequisite — Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold. - forall_elim · prerequisite — ∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds. - and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q. - mko_conn · cited — For propositions P and Q, ¬P denotes negation, P∧Q says that both hold, P∨Q says that at least one holds, P→Q says that if P then Q, and P↔Q denotes the biconditional requiring both directions. · mko-logical-connectives - truth_table · prerequisite — Classical truth conditions: ¬P is true ⇔ P is false; P∧Q is true ⇔ both are true; P∨Q is true ⇔ at least one is true (including both); P→Q is false ⇔ P is true and Q is false, and true otherwise; P↔Q is true ⇔ P→Q and Q→P are both true, that is, P and Q have the same truth value. - int_no_gap · prerequisite — No gaps in the integers: for integers k and m, k≤m or k≥m+1. ## Main derivation S01 · definitionBefore \neg P,\ P\land Q,\ P\lor Q,\ P\to Q,\ P\leftrightarrow Q After \text{truth value of each compound determined by those of}\ P,\ Q Conditions: none Rule: mko_conn · Uses: G1 Leaf basis: mko_conn + truth_table By the MKO, ¬P is negation, P∧Q requires both, P∨Q requires at least one, P→Q says if P then Q, and P↔Q requires both directions. In the classical two-valued context (G1) these definitions are truth conditions (truth_table): the truth of a compound depends only on the truth of its parts, not on any causal or temporal link between them. S02 · impliesBefore 4\mid 12,\ 6\mid 12 After \mathrm{T},\ \mathrm{T} Conditions: none Rule: divides_def · Uses: G2, G3 Leaf basis: divides_def + exists_intro + numeric_evaluation 12=4·3 with 3 an integer, so 4∣12 is true; 12=6·2, so 6∣12 is true. Both components are true. S03 · impliesBefore 4\mid 12\ \lor\ 6\mid 12 After \mathrm{T} Conditions: none Rule: truth_table · Uses: S01, S02 Leaf basis: truth_table By the truth condition for ∨, the disjunction is true when at least one part is true; here both are, so '4∣12 ∨ 6∣12' is true. Misreading ∨ as exclusive or (exactly one true) would wrongly make it false; the MKO's ∨ is inclusive and allows both to hold. S04 · impliesBefore \text{assume}\ 4\mid n\ (n\in\mathbb{Z}\ \text{arbitrary}) After n=4k\ \text{for some}\ k\in\mathbb{Z} Conditions: none Rule: divides_def · Uses: G2 Leaf basis: divides_def + exists_elim To prove ∀n (P(n)→Q(n)): take an arbitrary integer n with no extra assumption, and reason inside the temporary assumption 4∣n. By the definition of divisibility there is an integer k with n=4k (k is a fresh name for this witness). S05 · impliesBefore n=4k After n=2(2k),\ 2k\in\mathbb{Z},\ \text{so}\ 2\mid n Conditions: none Rule: mul_assoc · Uses: S04 Leaf basis: mul_assoc + divides_def + exists_intro + numeric_evaluation 4=2·2, so by associativity 4k=(2·2)k=2(2k); 2k is an integer (the integers are closed under multiplication), so 2∣n by the definition of divisibility. S06 · impliesBefore \text{under the assumption}\ 4\mid n:\ 2\mid n After \forall n\in\mathbb{Z}:\ P(n)\to Q(n) Conditions: none Rule: imp_intro · Uses: S04, S05 Leaf basis: imp_intro + forall_intro Under the temporary assumption 4∣n we derived 2∣n, so →-introduction discharges the assumption and gives P(n)→Q(n); n was arbitrary, so generalization gives ∀n∈ℤ (P(n)→Q(n)). Only one direction has been proved. S07 · impliesBefore n=2 After Q(2):\ 2=2\cdot 1,\ \mathrm{T} Conditions: none Rule: divides_def · Uses: G2 Leaf basis: divides_def + exists_intro + numeric_evaluation Now examine the reverse direction Q(n)→P(n). Take n=2: 2=2·1 with 1 an integer, so 2∣2 and Q(2) is true. S08 · case_splitBefore k\in\mathbb{Z}\ \text{with}\ 4k=2\ ? After k\le 0\ \lor\ k\ge 1 Conditions: k\in\mathbb{Z} Rule: int_no_gap · Uses: G2 Leaf basis: int_no_gap + case_analysis P(2) would need an integer k with 4k=2. By int_no_gap (m=0) every integer k satisfies k≤0 or k≥1; the two branches are handled in S08.1 and S08.2. S09 · impliesBefore Q(2)=\mathrm{T},\ \neg\exists k\in\mathbb{Z}:\ 4k=2 After P(2)=\mathrm{F};\ Q(2)\to P(2)=\mathrm{F};\ \neg\forall n\,(Q(n)\to P(n)) Conditions: none Rule: quantifier_negation · Uses: S07, S08, S08.1, S08.2 Leaf basis: case_analysis + truth_table + quantifier_negation Both branches give 4k≠2, so no integer k has 4k=2 and P(2) is false. By the truth condition for →, a true antecedent with a false consequent makes the implication false, so Q(2)→P(2) is false. n=2 is a counterexample, so by negation of quantifiers ∀n∈ℤ (Q(n)→P(n)) is false. S10 · impliesBefore \forall n\,(P(n)\to Q(n))\ \text{true},\ \forall n\,(Q(n)\to P(n))\ \text{false} After \forall n\in\mathbb{Z}:\ P(n)\leftrightarrow Q(n)\ \text{is false} Conditions: none Rule: contradiction · Uses: S06, S09 Leaf basis: contradiction + iff_intro + forall_elim Suppose ∀n (P(n)↔Q(n)) were true. Taking n=2 gives P(2)↔Q(2), and by the definition of the biconditional Q(2)→P(2), contradicting S09. So the universal biconditional is false. The direction proved in S06 holds for every n, but one direction is not enough for the biconditional. S11 · impliesBefore n=3 After P(3)=\mathrm{F},\ Q(3)=\mathrm{F} Conditions: none Rule: divides_def · Uses: G2 Leaf basis: divides_def + int_no_gap + order_basics Both components at n=3 are false: S11.1 shows no integer k has 4k=3, and S11.2 shows no integer k has 2k=3. S12 · impliesBefore P(3)=\mathrm{F},\ Q(3)=\mathrm{F} After P(3)\to Q(3)=\mathrm{T} Conditions: none Rule: truth_table · Uses: S11, S11.1, S11.2, S06 Leaf basis: truth_table + forall_elim By the truth condition for →, a false antecedent makes the implication true, so P(3)→Q(3) is true; this agrees with S06 instantiated at n=3 (∀-elimination). Although 2∤3, the implication is true: its truth depends only on truth values and does not mean that 4∣3 'causes' anything. S13 · impliesBefore 2+2=4,\ 7=2\cdot 3+1 After (2+2=4)\to(7\ \text{is odd})=\mathrm{T} Conditions: none Rule: truth_table · Uses: G3, S01 Leaf basis: truth_table + exists_intro + numeric_evaluation 2+2=4 is true; 7=2·3+1 with 3 an integer, so 7 is odd, which is true. By the truth condition for →, true→true is true. There is no causal or inferential link between the two statements, yet the implication is true: material implication is a truth condition, not causation and not temporal order. S14 · impliesBefore \neg Q(2)\land P(2)\quad\text{vs}\quad\neg(Q(2)\land P(2)) After \mathrm{F}\quad\text{vs}\quad\mathrm{T} Conditions: none Rule: truth_table · Uses: S07, S09 Leaf basis: truth_table Q(2) is true (S07) and P(2) is false (S09). In ¬Q(2)∧P(2), ¬ applies only to Q(2) and the main connective is ∧: ¬T∧F=F∧F=F. In ¬(Q(2)∧P(2)) the main connective is ¬: Q(2)∧P(2)=T∧F=F, whose negation is T. The brackets change the main connective and with it the truth value. S15 · impliesBefore \text{conclusions of S03, S06, S09, S10, S12, S13, S14} After \text{answers to parts (1) to (5)} Conditions: none Rule: and_intro_elim · Uses: S03, S06, S09, S10, S12, S13, S14 Leaf basis: and_intro_elim (1) '4∣12 ∨ 6∣12' is true (S03); (2) ∀n (P(n)→Q(n)) is proved (S06); (3) ∀n (Q(n)→P(n)) is false (S09), so ∀n (P(n)↔Q(n)) is false (S10); (4) P(3)→Q(3) and '2+2=4 → 7 is odd' are both true (S12, S13); (5) ¬Q(2)∧P(2) is false and ¬(Q(2)∧P(2)) is true (S14). ## Substeps and lemmas S08.1 · impliesBefore k\le 0 After 4k\le 0<2,\ \text{so}\ 4k\neq 2 Conditions: 4>0 Rule: order_basics · Uses: S08 Leaf basis: order_basics + mul_zero If k<0, multiplying by the positive number 4 gives 4k<0; if k=0 then 4k=0. So 4k≤0<2 and 4k≠2. S08.2 · impliesBefore k\ge 1 After 4k\ge 4>2,\ \text{so}\ 4k\neq 2 Conditions: 4>0 Rule: order_basics · Uses: S08 Leaf basis: order_basics + mul_one If k>1, multiplying by 4 gives 4k>4; if k=1 then 4k=4. So 4k≥4>2 and 4k≠2. S11.1 · impliesBefore 4k=3,\ k\in\mathbb{Z} After \text{impossible: } k\le 0\Rightarrow 4k\le 0<3;\ k\ge 1\Rightarrow 4k\ge 4>3 Conditions: 4>0 Rule: int_no_gap · Uses: S11 Leaf basis: int_no_gap + order_basics + case_analysis By int_no_gap, k≤0 or k≥1. If k≤0, multiplying by 4>0 gives 4k≤0<3; if k≥1, 4k≥4>3. Both branches give 4k≠3, so 4∤3 and P(3) is false. S11.2 · impliesBefore 2k=3,\ k\in\mathbb{Z} After \text{impossible: } k\le 1\Rightarrow 2k\le 2<3;\ k\ge 2\Rightarrow 2k\ge 4>3 Conditions: 2>0 Rule: int_no_gap · Uses: S11 Leaf basis: int_no_gap + order_basics + case_analysis By int_no_gap (m=1), k≤1 or k≥2. If k≤1 then 2k≤2<3; if k≥2 then 2k≥4>3. Both branches give 2k≠3, so 2∤3 and Q(3) is false. ## Reverse audits - S15 · sufficient_goal_reduction · S03, S06, S09, S10, S12, S13, S14, and_intro_elim — Goal G: answers to the five parts. Sufficient condition H: the conclusions of the listed steps; H⇒G just lists them. H is proved forwards from G1 to G3; G is never used as a premise. - S14 · premise_audit · S07, S09, truth_table — Both use the same component truth values; they differ only in structure. - S13 · premise_audit · truth_table, numeric_evaluation, exists_intro — Only the truth values of the two components are used. - S12 · premise_audit · truth_table, S06, forall_elim — Only the false antecedent is needed; Q(3) being false illustrates that the implication is true even with a false consequent. - S11.2 · premise_audit · int_no_gap, order_basics, case_analysis — The two branches cover every integer. - S11.1 · premise_audit · int_no_gap, order_basics, case_analysis — The two branches cover every integer. - S11 · premise_audit · divides_def, int_no_gap, order_basics — Both non-existence claims are proved in the sub-steps by case splits that cover all integers. - S10 · premise_audit · S06, S09, iff_intro, forall_elim, contradiction — It uses that a biconditional contains both directions (iff_intro); it does not claim P(n)↔Q(n) is false for every n: at n=4, 4=4·1 and 4=2·2, both parts are true and the biconditional is true. - S09 · premise_audit · S07, S08, case_analysis, truth_table, quantifier_negation — One counterexample suffices to refute the universal implication; case_analysis covered every integer k. - S08.2 · premise_audit · order_basics, mul_one — As in S08.1. - S08.1 · premise_audit · order_basics, mul_zero — 4>0 keeps the direction of the inequality; no converse is claimed. - S08 · premise_audit · int_no_gap — The two branches cover all integers; this relies on k being an integer (k=1/2 lies in neither). - S07 · premise_audit · divides_def, exists_intro — The single integer witness 1 suffices. - S06 · premise_audit · S04, S05, imp_intro, forall_intro — The n of S04 carried no extra condition, so generalization is allowed; the conclusion is P→Q only, and Q→P is not claimed. - S05 · premise_audit · mul_assoc, divides_def, exists_intro — 2k must be an integer to serve as the witness; N3 is an independent check of the identity 4k=2(2k). - S04 · premise_audit · divides_def, exists_elim, imp_intro — The temporary assumption is used only within S04 to S06 and is discharged in S06. - S03 · premise_audit · S02, truth_table — One true component is enough; both being true does not change the verdict. - S02 · premise_audit · divides_def, exists_intro, numeric_evaluation — Each divisibility needs one integer witness (3 and 2), and both can be recomputed (N1, N2). - S01 · premise_audit · G1, mko_conn, truth_table — This step only fixes the definitions and derives no new fact. ## Error controls and negative controls Disjunction is inclusiveThe MKO names the mistake of reading disjunction as exclusive or. 4∣12 and 6∣12 are both true, so '4∣12 ∨ 6∣12' is true (S03); reading ∨ as exclusive would wrongly make it false. Everyday 'or' is sometimes exclusive; the mathematical ∨ is not. Implication is not causationThe next mistake identified in the MKO is reading implication as causation. P(3)→Q(3) is true only because its antecedent is false (S12); '2+2=4 → 7 is odd' is true only because both parts are true (S13), with no causal link. Proving 'if P then Q' in every case still requires reasoning like S04 to S06, but the truth of a single implication depends only on truth values. A biconditional needs both directionsThe third mistake identified in the MKO is proving one direction and claiming the biconditional. S06 proves 4∣n → 2∣n for every integer, but n=2 breaks the converse (S09), so ∀n (4∣n ↔ 2∣n) is false (S10). Proving P↔Q requires proving P→Q and Q→P separately. - '4∣12 or 6∣12' is false because both hold → reject — ∨ is inclusive; a disjunction with both parts true is true (S03). - ∀n(4∣n→2∣n) is proved, so ∀n(4∣n↔2∣n) → reject — At n=2, 2∣2 is true and 4∣2 is false, so the converse fails (S09, S10). - '4∣3 → 2∣3' is false because 2∤3 → reject — The antecedent 4∣3 is false, so the implication is true (S12). - ¬A∧B and ¬(A∧B) mean the same → reject — Take A to be 2∣2 (true) and B to be 4∣2 (false): the first is false and the second true (S14). ## Result and limitations '4∣12 ∨ 6∣12' is true, because ∨ is inclusive. ∀n∈ℤ (4∣n → 2∣n) holds, but ∀n∈ℤ (2∣n → 4∣n) is false (counterexample n=2), so ∀n∈ℤ (4∣n ↔ 2∣n) is false. 4∣3 → 2∣3 and '2+2=4 → 7 is odd' are both true; implication depends only on truth values and does not express causation. ¬(2∣2)∧(4∣2) is false and ¬((2∣2)∧(4∣2)) is true; the brackets fix the main connective. - The truth conditions are definitions for the classical two-valued context; in intuitionistic logic → and ∨ have different meanings, which this guide does not treat. - N3 is a degree-one polynomial identity check on a grid and only cross-checks 4k=2(2k) in S05; the generality comes from the reasoning of S04 to S06. ## Verification and source binding 13 declared checks for this guide; only the listed expressions are covered. - source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f - adopted source SHA-256: 8c4378aa84c06b4b7dd718820bdb2b5a2f72f9a464ab78ec19745a2813a16d7a - original source SHA-256: 8c4378aa84c06b4b7dd718820bdb2b5a2f72f9a464ab78ec19745a2813a16d7a - Lean: NOT_RUN · independent AI: NOT_MEASURED [Same-source full JSON (formulas, ASTs, controls and revision log)](https://ocme.evemisslab.com/data/derivations/production-20261019/guides/derivation-mko-logical-connectives-01.json)