Logical connectives: testing ∨, →, ↔ and brackets on divisibility statements

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 u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw.
• 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

\neg P,\ P\land Q,\ P\lor Q,\ P\to Q,\ P\leftrightarrow Q

\text{truth value of each compound determined by those of}\ P,\ Q

4\mid 12,\ 6\mid 12

\mathrm{T},\ \mathrm{T}

4\mid 12\ \lor\ 6\mid 12

\mathrm{T}

\text{assume}\ 4\mid n\ (n\in\mathbb{Z}\ \text{arbitrary})

n=4k\ \text{for some}\ k\in\mathbb{Z}

n=4k

n=2(2k),\ 2k\in\mathbb{Z},\ \text{so}\ 2\mid n

\text{under the assumption}\ 4\mid n:\ 2\mid n

\forall n\in\mathbb{Z}:\ P(n)\to Q(n)

n=2

Q(2):\ 2=2\cdot 1,\ \mathrm{T}

k\in\mathbb{Z}\ \text{with}\ 4k=2\ ?

k\le 0\ \lor\ k\ge 1

Q(2)=\mathrm{T},\ \neg\exists k\in\mathbb{Z}:\ 4k=2

P(2)=\mathrm{F};\ Q(2)\to P(2)=\mathrm{F};\ \neg\forall n\,(Q(n)\to P(n))

\forall n\,(P(n)\to Q(n))\ \text{true},\ \forall n\,(Q(n)\to P(n))\ \text{false}

\forall n\in\mathbb{Z}:\ P(n)\leftrightarrow Q(n)\ \text{is false}

n=3

P(3)=\mathrm{F},\ Q(3)=\mathrm{F}

P(3)=\mathrm{F},\ Q(3)=\mathrm{F}

P(3)\to Q(3)=\mathrm{T}

2+2=4,\ 7=2\cdot 3+1

(2+2=4)\to(7\ \text{is odd})=\mathrm{T}

\neg Q(2)\land P(2)\quad\text{vs}\quad\neg(Q(2)\land P(2))

\mathrm{F}\quad\text{vs}\quad\mathrm{T}

\text{conclusions of S03, S06, S09, S10, S12, S13, S14}

\text{answers to parts (1) to (5)}

Substeps and lemmas

k\le 0

4k\le 0<2,\ \text{so}\ 4k\neq 2

k\ge 1

4k\ge 4>2,\ \text{so}\ 4k\neq 2

4k=3,\ k\in\mathbb{Z}

\text{impossible: } k\le 0\Rightarrow 4k\le 0<3;\ k\ge 1\Rightarrow 4k\ge 4>3

2k=3,\ k\in\mathbb{Z}

\text{impossible: } k\le 1\Rightarrow 2k\le 2<3;\ k\ge 2\Rightarrow 2k\ge 4>3

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 inclusive

The 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 causation

The 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 directions

The 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

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