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Implicational propositional calculus

Implicational propositional calculus is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Implicational propositional calculus rather than just read about it. In short: In mathematical logic, the implicational propositional calculus is a version of classical propositional calculus that uses only one connective, called implication or conditional. In formulas, this binary operation is indicated by "implies", "if ..., then ...", "→", " → {\displaystyle \rightarrow } ", etc..

Key takeaways

  • Implicational propositional calculus belongs to mathematics; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Implicational propositional calculus to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Implicational propositional calculus from memory before moving on to harder problems.

Reference excerpt

In mathematical logic, the implicational propositional calculus is a version of classical propositional calculus that uses only one connective, called implication or conditional. In formulas, this binary operation is indicated by "implies", "if ..., then ...", "→", " → {\displaystyle \rightarrow } ", etc..

Functional (in)completeness Implication alone is not functionally complete as a logical operator because one cannot form all other two-valued truth functions from it. For example, the two-place truth function that always returns false is not definable from → and arbitrary propositional variables: any formula constructed from → and propositional variables must receive the value true when all of its variables are evaluated to true. It follows that {→} is not functionally complete. However, if one adds a nullary connective ⊥ for falsity, then one can define all other truth functions. Formulas over the resulting set of connectives {→, ⊥} are called f-implicational. If P and Q are propositions, then:

¬P is equivalent to P → ⊥ P ∧ Q is equivalent to (P → (Q → ⊥)) → ⊥ P ∨ Q is equivalent to (P → Q) → Q P ↔ Q is equivalent to ((P → Q) → ((Q → P) → ⊥)) → ⊥ Since the above operators are known to be functionally complete, it follows that any truth function can be expressed in terms of → and ⊥.

Axiom system The following statements are considered tautologies (irreducible and intuitively true, by definition).

Axiom schema 1 is P → (Q → P). Axiom schema 2 is (P → (Q → R)) → ((P → Q) → (P → R)). Axiom schema 3 (Peirce's law) is ((P → Q) → P) → P. The one non-nullary rule of inference (modus ponens) is: from P and P → Q infer Q. Where in each case, P, Q, and R may be replaced by any formulas that contain only "→" as a connective. If Γ {\displaystyle \Gamma } is a set of formulas and A a formula, then Γ ⊢ A {\displaystyle \Gamma \vdash A} means that A is derivable using the axioms and rules above and formulas from Γ {\displaystyle \Gamma } as additional hypotheses. Łukasiewicz (1948) found an axiom system for the implicational calculus that replaces the schemas 1–3 above with a single schema

((P → Q) → R) → ((R → P) → (S → P)). He also argued that there is no shorter axiom system.

Basic properties of derivation Since all axioms and rules of the calculus are schemata, derivation is closed under substitution:

If Γ ⊢ A , {\displaystyle \Gamma \vdash A,} then σ ( Γ ) ⊢ σ ( A ) , {\displaystyle \sigma (\Gamma )\vdash \sigma (A),}

where σ is any substitution (of formulas using only implication). The implicational propositional calculus also satisfies the deduction theorem:

If Γ , A ⊢ B {\displaystyle \Gamma ,A\vdash B} , then Γ ⊢ A → B . {\displaystyle \Gamma \vdash A\to B.}

As explained in the deduction theorem article, this holds for any axiomatic extension of the system containing axiom schemas 1 and 2 above and modus ponens.

Completeness The implicational propositional calculus is semantically complete with respect to the usual two-valued semantics of classical propositional logic. That is, if Γ is a set of implicational formulas, and A is an implicational formula entailed by Γ, then Γ ⊢ A {\displaystyle \Gamma \vdash A} .

Proof A proof of the completeness theorem is outlined below. First, using the compactness theorem and the deduction theorem, we may reduce the completeness theorem to its special case with empty Γ, i.e., we only need to show that every tautology is derivable in the system. The proof is similar to completeness of full propositional logic, but it also uses the following idea to overcome the functional incompleteness of implication. If A and F are formulas, then A → F is equivalent to (¬A*) ∨ F, where A* is the result of replacing in A all, some, or none of the occurrences of F by falsity. Similarly, (A → F) → F is equivalent to A* ∨ F. So under some conditions, one can use them as substitutes for saying A* is false or A* is true respectively. We first observe some basic facts about derivability:

Indeed, we can derive A → (B → C) using Axiom 1, and then derive A → C by modus ponens (twice) from Ax. 2.

This follows from (1) by the deduction theorem.

If we further assume C → B, we can derive A → B using (1), then we derive C by modus ponens. This shows A → C , ( A → B ) → C , C → B ⊢ C {\displaystyle A\to C,(A\to B)\to C,C\to B\vdash C} , and the deduction theorem gives A → C , ( A → B ) → C ⊢ ( C → B ) → C {\displaystyle A\to C,(A\to B)\to C\vdash (C\to B)\to C} . We apply Ax. 3 to obtain (3). Let F be an arbitrary fixed formula. For any formula A, we define A0 = (A → F) and A1 = ((A → F) → F). Consider only formulas in propositional variables p1, ..., pn. We claim that for every formula A in these variables and every truth assignment e,

We prove (4) by induction on A. The base case A = pi is trivial. Let A = (B → C). We distinguish three cases:

e(C) = 1. Then also e(A) = 1. We have

( C → F ) → F ⊢ ( ( B → C ) → F ) → F {\displaystyle (C\to F)\to F\vdash ((B\to C)\to F)\to F}

by applying (2) twice to the axiom C → (B → C). Since we have derived (C → F) → F by the induction hypothesis, we can infer ((B → C) → F) → F. e(B) = 0. Then again e(A) = 1. The deduction theorem applied to (3) gives

… excerpt ends here. Continue reading the full article.

Worked examples

Example 1 — a first encounter with Implicational propositional calculus

Start with the simplest possible case. Write down what Implicational propositional calculus claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Implicational propositional calculus before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Implicational propositional calculus ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Implicational propositional calculus

In research
Implicational propositional calculus appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Implicational propositional calculus in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Implicational propositional calculus is common in secondary-school and first-year university syllabi. It links to neighbouring topics Conditionals, Propositional calculus, Systems of formal logic, so understanding it makes those chapters shorter.
In everyday life
Look for Implicational propositional calculus outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

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How to study Implicational propositional calculus in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Implicational propositional calculus means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Implicational propositional calculus out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Implicational propositional calculus in simple terms?

In mathematical logic, the implicational propositional calculus is a version of classical propositional calculus that uses only one connective, called implication or conditional. In formulas, this binary operation is indicated by "implies", "if ..., then ...", "→", " → {\displaystyle \rightarrow }…

Why does Implicational propositional calculus matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Implicational propositional calculus?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Implicational propositional calculus.

Tags

  • Conditionals
  • Propositional calculus
  • Systems of formal logic

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