PDDL-Prover 1.0.0

A python prover for evaluating FOL formulas on PDDL

PDDL-Prover

A python prover for evaluating FOL formulas on PDDL.

PDDL-Prover allows to construct arbitrary formulas in first-order logic (FOL) in Python, using a declarative syntax and including counting quantifiers. Then, the truth values of these formulas can be evaluated on a knowledge base, which represents a set of objects/contants and a set of true atoms (as in PDDL).

This results useful for evaluating conditions on PDDL states. For example, to check if in a given blocksworld state block A is on top of the table and under block B, we can do the following:

(ontable(A) & on(B,A)).evaluate(kb)

where kb is the knowledge base containing the objects and atoms of the blocksworld state. The line above will return True if the condition is met at the state, and False otherwise. You can find a more elaborate example below.

Installation

The package is available in PyPI and can be installed with pip install pddl-prover. Alternatively, you can clone this repository.

Example of use

from pddl-prover import *

b0, b1, b2, b3, b4  =  Constant(0), Constant(1), Constant(2), Constant(3), Constant(4)
x, y  =  Variable('x'), Variable('y')

handempty  =  Predicate('handempty', 0) # The second argument is the arity
ontable  =  Predicate('ontable', 1)
on  =  Predicate('on', 2)
clear  =  Predicate('clear', 1)

# Note: atoms in the knowledge base can be represented either in tuple form or as instances of the class logic.Atom
kb  = ({b0,b1,b2},{('ontable', (0,)), ('on',(1,0)), ('on', (2,1)), ('clear',(2,)),
('handempty', tuple())})

print("Evaluation:", handempty().evaluate(kb)) # True
print("Evaluation:", (ontable(b0) & on(b1,b0)).evaluate(kb)) # True
print("Evaluation:", TE(x, on(x,b0)).evaluate(kb)) # There Exists some x for which on(x,b0) -> True
print("Evaluation:", (TE(x, ontable(x)) == 1).evaluate(kb)) # There exists <only> one object for which ontable(x) is True -> True (counting quantifier)
print("Number of true substitutions:", Count(on(x,y), x, y).evaluate(kb)[0]) # Returns the number of x and y substitutions for which on(x,y) is True -> 2