truth-value semantics for intuitionistic propositional logic is sound


Proof.

We show that, for each positive integer n, every theorem of intuitionistic propositional logicPlanetmathPlanetmath is a tautologyMathworldPlanetmath for Vn. This amounts to showing that, for every interpretationMathworldPlanetmathPlanetmath v on Vn,

  • •

    each axiom is true, and

  • •

    modus ponensMathworldPlanetmath preserves truth.

Let us take care of the second one first. Suppose v(A)=v(A→B)=n. If v⁢(A)≤v⁢(B), then v⁢(B)=n. Otherwise, v⁢(B)<v⁢(A). But this means that n=v(A→B)=v(B), forcingMathworldPlanetmath v⁢(B)=n. Therefore, v⁢(B)=n.

Now, we verify that each of the axiom schemasMathworldPlanetmath below are true:

  1. 1.

    (A∧B)→A and (A∧B)→B.

    Since v⁢(A∧B)=min⁡{v⁢(A),v⁢(B)}≤v⁢(A), we get v((A∧B)→A)=n. The other one is proved similarly.

  2. 2.

    A→(A∨B) and B→(A∨B).

    Since v⁢(A)≤max⁡{v⁢(A),v⁢(B)}=v⁢(A∨B), we get v(A→(A∨B))=n. The other one is proved similarly.

  3. 3.

    A→(B→A).

    If v⁢(B)≤v⁢(A), v(B→A)=n, so that v(A→(B→A))=n as well. If v⁢(A)<v⁢(B), then v(B→A)=v(A), so that v(A→(B→A))=n.

  4. 4.

    ¬A→(A→B).

    If v⁢(A)≤v⁢(B), v(A→B)=n, so that v(¬A→(A→B))=n as well. If v⁢(B)<v⁢(A), then v(A→B)=v(B). Also, v⁢(B)<v⁢(A) implies that v⁢(A)>0, so that v⁢(¬⁢A)=0, and v(¬A→(A→B))=n as a result.

  5. 5.

    A→(B→(A∧B)).

    If v⁢(B)=v⁢(A∧B), then v⁢(B)≤v⁢(A) and v(B→(A∧B))=n, so that v(A→(B→(A∧B))=n also. If on the other hand v⁢(A∧B)<v⁢(B), then

    v(A)=v(A∧B) and v(B→(A∧B))=v(A∧B),

    so that

    v(A→(B→(A∧B)))=v(A→(A∧B))=n.
  6. 6.

    (A→C)→((B→C)→((A∨B)→C)).

    If v⁢(B)=v⁢(A∨B), then v⁢(A)≤v⁢(B), and

    v((B→C)→((A∨B)→C))=v((B→C)→(B→C))=n,

    so that

    v((A→C)→((B→C)→((A∨B)→C)))=n

    as well. Otherwise, v⁢(B)<v⁢(A)=v⁢(A∨B). This means that

    v((B→C)→((A∨B)→C)))=v((B→C)→(A→C)),

    and therefore

    v((A→C)→((B→C)→((A∨B)→C)))=v((A→C)→((B→C)→(A→C)))=n

    by 3 above.

  7. 7.

    (A→B)→((A→(B→C))→(A→C)).

    It is clear that v(C)≤v(B→C). If v(C)=v(B→C), then

    v((A→(B→C))→(A→C))=v((A→C)→(A→C))=n,

    so that

    v((A→B)→((A→(B→C))→(A→C)))=n

    too. If v(C)<v(B→C), then v(B→C)=n, which implies v⁢(B)≤v⁢(C). This in turn implies that v(A→B)≤v(A→C), so that

    v((A→C)→((A→(B→C))→(A→C)))≤v((A→B)→((A→(B→C))→(A→C))).

    But by 3 above,

    v((A→C)→((A→(B→C))→(A→C)))=n,

    hence

    v((A→B)→((A→(B→C))→(A→C)))=n

    as a result.

  8. 8.

    (A→B)→((A→¬B)→¬A).

    Pick any C such that v⁢(C)=0, such as D∧¬⁢D. Then v(¬B)=v(B→C), so that

    v((A→B)→((A→¬B)→¬A))=v((A→B)→((A→(B→C))→(A→C)))=n

    by 7.

∎

Note that the proofs of the axioms employ some elementary facts, for any wff’s A,B,C:

  • •

    If v⁢(B)=n or v⁢(A)=0, then v(A→B)=n.

  • •

    if v⁢(B)=0, then v(A→B)=v(¬A).

  • •

    if v⁢(A)=n, then v(A→B)=v(B).

  • •

    v(B)≤v(A→B).

  • •

    if v⁢(B)≤v⁢(C), then

    • –

      v⁢(A∨B)≤v⁢(A∨C),

    • –

      v⁢(A∧B)≤v⁢(A∧C),

    • –

      v(A→B)≤v(A→C),

    • –

      v(C→A)≤v(B→A).

From the facts above, one readily deduces:

  • •

    if v⁢(B)≤v⁢(C), then v⁢(¬⁢C)≤v⁢(¬⁢B),

  • •

    if v⁢(B)=v⁢(C), then

    • –

      v⁢(A∨B)=v⁢(A∨C),

    • –

      v⁢(A∧B)=v⁢(A∧C),

    • –

      v(A→B)=v(A→C),

    • –

      v(C→A)=v(B→A),

    • –

      v⁢(¬⁢B)=v⁢(¬⁢C).

Title truth-value semantics for intuitionistic propositional logic is sound
Canonical name TruthvalueSemanticsForIntuitionisticPropositionalLogicIsSound
Date of creation 2013-03-22 19:31:38
Last modified on 2013-03-22 19:31:38
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 21
Author CWoo (3771)
Entry type Definition
Classification msc 03B20