# Reasoning in LTN¶

This tutorial defines and illustrates reasoning in LTN. It expects basic familiarity with other parts of LTN.

### Logical Consequence in LTN¶

The essence of reasoning is to determine if a closed formula $\phi$ is the logical consequence of a knowledgebase $(\mathcal{K},\mathcal{G}_\theta,\Theta)$, where $\mathcal{K}$ denotes the set of rules in the knowledgebase and $\mathcal{G}_\theta$ denotes a grounding that depends on some parameters $\theta \in \Theta$.

The notion of logical consequence is adapted to Real Logic as follows:

• In classical logic (boolean truth values), a formula $\phi$ is the logical consequence of a knowledgebase $\mathcal{K}$ if, for every interpretation (or model) that verifies the formulas in $\mathcal{K}$, $\phi$ is verified;
• In Real Logic (fuzzy truth values), a formula $\phi$ is the logical consequence of $(\mathcal{K},\mathcal{G}_\theta,\Theta)$ if, for every grounding $\mathcal{G}_\theta$ such that $\mathrm{SatAgg}_{\phi'\in\mathcal{K}}\mathcal{G}_{\theta}(\phi') \geq q$, we have $\mathcal{G}_\theta(\phi)\geq q$, where $q$ is a fixed satisfaction threshold.

Logical consequence in Real Logic, by direct application of the definition, requires querying the truth value of $\phi$ for a potentially infinite set of groundings. We therefore, in practice, consider the following directions:

1. Reasoning by brave inference: one seeks to verify if for all the grounded theories that maximally satisfy $\mathcal{K}$, the grounding of $\phi$ gives a truth value greater than a threshold $q$. This often requires to check an infinite number of groundings. Instead, one can approximate the search for these grounded theories by running the optimization w.r.t. the knowledgebase satisfiability multiple times and checking these groundings only.
2. Reasoning by refutation: one seeks to find out a counterexample of a grounding that does satisfy the knowledgebase $\mathcal{K}$ but not the formula $\phi$ (given the threshold $q$). A directed search for such examples is performed using a different learning objective.

In this tutorial, we illustrate the second option, reasoning by refutation.

In [1]:
import logictensornetworks as ltn
import tensorflow as tf
import numpy as np
import matplotlib.pyplot as plt

### Example¶

We illustrate reasoning on the following toy example: $$(A \lor B) \models_q A \ ?$$ where $A$ and $B$ are two propositional variables, and $\frac{1}{2} < q < 1$ is the satisfaction threshold.

We define $\mathcal{K}=\{A \lor B\}$ and $\phi=A$.

In [2]:
A = ltn.proposition(0.,trainable=True)
B = ltn.proposition(0.,trainable=True)

Or = ltn.Wrapper_Connective(ltn.fuzzy_ops.Or_ProbSum())

def axioms():
return Or(A,B)

def phi():
return A

### Reasoning by Refutation¶

The goal is to find a grounding that satisfies $\mathcal{K}$ but does not satisfy $\phi$. One can perform a directed search for such a counterexample by minimizing $\mathcal{G}_\theta(\phi)$ while imposing a constraint that invalidates results where $\mathcal{G}_\theta(\mathcal{K})<q$.

Let us define $\mathrm{penalty}(\mathcal{G}_\theta,q)=\begin{cases} c \ \text{if}\ \mathcal{G}_\theta(\mathcal{K}) < q,\\ 0 \ \text{otherwise}, \end{cases}$ where $c>1$ and set the objective: $$\mathcal{G}^\ast = \mathrm{argmin}_{\mathcal{G}_\theta} (\mathcal{G}_\theta(\phi) + \mathrm{penalty}(\mathcal{G}_\theta,q))$$

The penalty $c$ ($>1$) is higher than any potential reduction in $\mathcal{G}(\phi)$ ($\leq 1$). $\mathcal{G}^\ast$ should satisfy in priority $\mathcal{G}^*(\mathcal{K}) \geq q$ before reducing $\mathcal{G}^*(\phi)$.

• If $\mathcal{G}^\ast(\mathcal{K}) < q$ : Then for all $\mathcal{G}_\theta$, $\mathcal{G}_\theta(\mathcal{K}) < q$ and therefore $(\mathcal{K},\mathcal{G}(\ \cdot\mid \theta), \Theta)\models_q\phi$.
• If $\mathcal{G}^\ast(\mathcal{K}) \geq q \ \text{and}\ \mathcal{G}^\ast(\phi) \geq q$ : Then for all $\mathcal{G}_\theta$ with $\mathcal{G}_\theta(\mathcal{K}) \geq q$, we have that $\mathcal{G}_\theta(\phi) \geq \mathcal{G}^\ast(\phi) \geq q$ and therefore $(\mathcal{K},\mathcal{G}(\ \cdot\mid\theta),\Theta)\models_q\phi$.
• If $\mathcal{G}^\ast(\mathcal{K}) \geq q \ \text{and}\ \mathcal{G}^\ast(\phi) < q$ : Then $(\mathcal{K},\mathcal{G}(\ \cdot\mid\theta),\Theta) \nvDash_q\phi$.

### Soft constraint¶

However, as $\mathrm{penalty}(\mathcal{G}_\theta,q)$ is a constant function on the continuous parts of its domain (zero gradients), it cannot be used directly as an objective to reach via gradient descent optimization.Instead, one should approximate the penalty with a soft constraint.

We use $\mathtt{elu}(\alpha,\beta (q-\mathcal{G}_\theta(\mathcal{K})))=\begin{cases} \beta (q-\mathcal{G}_\theta(\mathcal{K}))\ &\text{if}\ \mathcal{G}_\theta(\mathcal{K}) \leq 0,\\ \alpha (e^{q-\mathcal{G}_\theta(\mathcal{K})}-1) \ &\text{otherwise}, \end{cases}$ where $\alpha \geq 0$ and $\beta \geq 0$ are two hyper-parameters:

• When $\mathcal{G}_\theta(\mathcal{K}) < q$, the penalty is linear in $(q-\mathcal{G}_\theta(\mathcal{K}))$ with a slope of $\beta$. Setting $\beta$ high, the gradients for $\mathcal{G}_\theta(\mathcal{K})$ will be high in absolute value if the knowledgebase in not satisfied; the minimizer will prioritize increasing the satisfaction of the knowledgebase.
• When $\mathcal{G}_\theta(\mathcal{K}) > q$, the penalty is a negative exponential that converges to $-\alpha$. Setting $\alpha$ low but non zero ensures that, while the penalty plays an insignificant role when the knowledgebase is satisfied, the gradients do not vanish.
In [3]:
trainable_variables = [A,B]

# hyperparameters of the soft constraint
alpha = 0.05
beta = 10
# satisfaction threshold
q = 0.95

for epoch in range(4000):
sat_KB = axioms()
sat_phi = phi()
penalty = tf.keras.activations.elu(beta*(q-sat_KB),alpha=alpha)
loss = sat_phi + penalty
if epoch%400 == 0:
print("Epoch %d: Sat Level Knowledgebase %.3f Sat Level phi %.3f"%(epoch, axioms(), phi()))
print("Training finished at Epoch %d with Sat Level Knowledgebase %.3f Sat Level phi %.3f"%(epoch, axioms(), phi()))
Epoch 0: Sat Level Knowledgebase 0.002 Sat Level phi 0.001
Epoch 400: Sat Level Knowledgebase 0.591 Sat Level phi 0.358
Epoch 800: Sat Level Knowledgebase 0.862 Sat Level phi 0.610
Epoch 1200: Sat Level Knowledgebase 0.950 Sat Level phi 0.705
Epoch 1600: Sat Level Knowledgebase 0.950 Sat Level phi 0.615
Epoch 2000: Sat Level Knowledgebase 0.951 Sat Level phi 0.485
Epoch 2400: Sat Level Knowledgebase 0.962 Sat Level phi 0.319
Epoch 2800: Sat Level Knowledgebase 0.998 Sat Level phi 0.116
Epoch 3200: Sat Level Knowledgebase 1.000 Sat Level phi 0.000
Epoch 3600: Sat Level Knowledgebase 1.000 Sat Level phi 0.000
Training finished at Epoch 3999 with Sat Level Knowledgebase 1.000 Sat Level phi 0.000

At the end of training, the optimizer has found a grounding that satisfies $A \lor B$ but not $A$ (given the satisfaction threshold $q=0.95$). This is a counterexample to the logical consequence, proving that $A \lor B \nvDash A$