WOLFRAM|DEMONSTRATIONS PROJECT

Equivalential Calculus

all theorems

explain individual step

part of original

step

1. EEpqEErqEpr

2. EEsEErqEprEEpqs

3. EEpqEpq

4. EErEpqEEpqr

5. EEErqEprEpq

6. Epp

7. EEpqEqp

8. EErEqpEEpqr

9. EEEpqrErEqp

10. EErqEEprEpq

11. EEpEpqq

12. EqEpEpq

13. EErqEEpEpqr

14. EEpEqrEEpqr

15. EEEpqrEpEqr

16. EEpEqrEpErq

17. EEEprEqpErq

18. EEEpqrEEqpr

19. EEpEEqrsEpEErqs

20. EEpEEqrsEpEqErs

21. EEpEqErsEpEEqrs

Axiom

1{p  Epq,q  EErqEpr,r  s}, 1

2{s  Epq}, 1

1{p  Epq,q  Epq}, 3

4{r  Epq,p  Erq,q  Epr}, 1

5{r  p,q  p}, 3{q  p}

1{p  q,r  p}, 6{p  q}

1{p  Epq,q  Eqp}, 7

7{p  ErEqp,q  EEpqr}, 8

2{s  EEprEpq,r  p,p  r}, 9{q  r,r  Epq}

5{r  Epq,p  EpEpq}, 10{r  Epq}

7{p  EpEpq}, 11

1{p  EpEpq}, 11

2{s  EEpqr,r  q,q  Eqr}, 13{r  Epq,q  r,p  q}

7{p  EpEqr,q  EEpqr}, 14

9{p  Erq,q  p,r  EpEqr}, 9{p  r,r  p}

16{p  EEprEqp}, 5{r  p,q  r,p  q}

16{p  EEpqr,q  r,r  Eqp}, 9

10{r  EEqrs,q  EErqs}, 18{p  q,q  r,r  s}

10{r  EEqrs,q  EqErs}, 15{p  q,q  r,r  s}

7{p  EpEEqrs,q  EpEqErs}, 20

The equivalential calculus is a subsystem of propositional calculus with equivalence (≡) as the only connective. Two equivalential tautologies are

p≡p

and

(p≡q)≡(q≡p)

. In Polish notation, discovered by Łukasiewicz, these formulas are written as

Epp

and

EEpqEqp

. This Demonstration shows the derivation of 21 theorems of the equivalential calculus based on one axiom,

EEpqEErqEpr

, and the rules of substitution and modus ponens.

Here is an example of substitution. The substitution

{pEpq,qEEpqs}

means that the propositional variable

is replaced by

Epq

and that

is replaced by

EEpqs

. So applying that substitution to the formula

EpEpq

gives

EEpqEEpqEEpqs

Modus ponens is a derivation of

from

and

Exy

To be a theorem of the equivalential calculus, its formula must follow from already proven theorems. The first theorem is the axiom, but the rest of the numbered formulas need explanation.

EEpqEErqEpr

Axiom

EEsEErqEprEEpqs

1{pEpq,qEErqEpr,rs},1

The second line means that the formula

EEsEErqEprEEpqs

follows by modus ponens from formula 1 (the axiom), to which the substitution

S={pEpq,qEErqErp,rs}

is applied, and from the formula 1, which is the axiom.

gives

EE(Epq)(EErqEpr)EEs(EErqEpr)E(Epq)s

, which is

EEpqEErqEpr≡EEsEErqEprEEpqs

The formula

EEsEErqEprEEpqs

in step 2 follows from the last two by modus ponens.