Neuro-symbolic AI: Worked examples — AI-ML deep dive

Before we compute anything, we must be sure every symbol is earned. Several live here.

Definition Data loss

$L_{data}$ and the symbol $ℓ$ $L_{data}$ is the ordinary supervised loss — the number that measures how far the network's predictions $f_{θ} (x_{i})$ sit from the true labels $y_{i}$ on your training set. It comes straight from the parent note's $L_{data} = \frac{1}{N} \sum_{i} ℓ (f_{θ} (x_{i}), y_{i})$ . What is $ℓ$ ? $ℓ (prediction, label)$ is the per-example mismatch score — one number per training sample saying "how wrong was this single guess?" For classification it is usually cross-entropy (big when the network is confidently wrong), for regression usually mean squared error $(pred - label)^{2}$ . $L_{data}$ just averages $ℓ$ over all $N$ examples. On this page we never recompute $L_{data}$ ; we only add a symbolic penalty to it, forming $L_{total} = L_{data} + L_{symbolic} .$ Think of $L_{data}$ as "fit the examples" and $L_{symbolic}$ as "obey the rules"; training minimises their sum.

Now the machine. A handful of continuous operators do all the work; we define them on the light-slider $μ$ .

Formula The operators we will reuse everywhere

Soft AND (product t-norm): $a \land b \approx a \cdot b$ . Multiplying two sliders both between $0$ and $1$ gives something smaller than either — exactly what "both must hold" should feel like: the joint is never more certain than its weakest part. Soft NOT: $\neg a = 1 - a$ . Flip the slider. Clamp / positive-part $[\cdot]_{+}$ — the penalty floor. Define $[x]_{+} = max (0, x) = {x 0 x > 0 x \leq 0 .$ Read $[x]_{+}$ as "keep $x$ if it is positive, otherwise floor it to $0$ ." Its only job is "never let a penalty go negative." From here on the violation of an implication is written with this dedicated symbol: $violation (a \Rightarrow b) = [a - b]_{+} .$ Why floor at $0$ and not just $a - b$ ? If $b \geq a$ the rule is already satisfied — we do not want to reward over-satisfaction with a negative penalty, that would push $b$ artificially high. The $[\cdot]_{+}$ floor says "once you obey, stop pushing." That floor is the source of all the case-splitting on this page. Merge / OR of truth degrees — a genuinely different operation, written $\lor$ . When the same fact is supported by two independent derivations with degrees $u$ and $v$ , we keep the strongest: $u \lor v = max (u, v) .$ This is the fuzzy "OR": "the fact is at least as true as its best evidence." We use it to fold a newly derived degree into whatever we already believed. Do not confuse $[\cdot]_{+}$ with $\lor$ . They are written with different symbols on this page precisely so you never mix them: $[\cdot]_{+}$ is a one-sided floor on a penalty (one input is the constant $0$ ); $\lor$ is a two-input best-evidence combiner over truth degrees. They happen to share the same underlying $max$ but do opposite jobs.

The scenario matrix

Every worked example below is tagged with the matrix cell it fills. The matrix is organised by the sign / magnitude of the violation the rule produces, plus the special inputs a logic engine can hit.

Cell	What makes it special	Where it bites	Example
A Violation $> 0$	rule broken, gradient nonzero	constraint pushes network	Ex 1
B Violation $= 0$ (satisfied)	floor active, gradient dies	why training goes quiet	Ex 2
C Violation $= 0$ (over-satisfied)	$b > a$ , floor masks the slack	subtle no-op case	Ex 2
D Chained soft-AND, many terms	products shrink toward $0$	vanishing-confidence trap	Ex 3
E Mutual-exclusivity (two-sided)	both classes fight for mass	one-hot enforcement	Ex 4
F Degenerate fact $μ = 0$	premise unknown/false	inference produces nothing	Ex 5
G Certain fact $μ = 1$	premise fully true	rule fires cleanly	Ex 5
H Fixed-point / limiting behaviour	forward chaining converges	when to stop iterating	Ex 6
I Real-world word problem	end-to-end pipeline	loan-approval story	Ex 7
J Exam twist	sign trap + weighting $λ$	the "gotcha"	Ex 8

Prerequisites this page leans on: 6.1.03-Neural-Network-Architectures (what $f_{θ}$ and backprop are), 6.2.04-Knowledge-Graphs (predicates like $parent (X, Y)$ ), and 6.4.02-Explainable-AI (why interpretable outputs matter). Related frontiers: 6.5.10-Causal-Inference, 6.5.01-Few-Shot-Learning.

Worked examples

Worked example Ex 1 — Cell A: a live violation (positive penalty)

Statement. A chest-X-ray network outputs $p (pneumonia_{L}) = 0.80$ and $p (opacity_{L}) = 0.30$ (recall $p \equiv μ$ ). The anatomical rule is "pneumonia in the left lung $\Rightarrow$ high left-lung opacity". Constraint weight $λ = 2.0$ . Find the symbolic penalty added to the loss, and say which way each probability is pushed.

Forecast: guess — is the penalty positive or zero? Which number moves down, which moves up?

Identify $a$ and $b$ . Here $a = p (pneumonia_{L}) = 0.80$ (the "if" part), $b = p (opacity_{L}) = 0.30$ (the "then" part). Why this step? Implication violation is always measured as (premise) minus (conclusion); we must name which is which before subtracting.
Compute raw gap. $a - b = 0.80 - 0.30 = 0.50$ . Why this step? The gap is how much the "if" outruns the "then" — the raw evidence of a broken rule.
Apply the floor $[\cdot]_{+}$ . $violation = [0.50]_{+} = 0.50$ . Positive, so we are in Cell A. Why this step? The floor lets us skip penalising already-obeyed rules; here it does nothing because the gap is positive.
Weight and add. $L_{symbolic} = λ \cdot 0.50 = 2.0 \times 0.50 = 1.00$ , so $L_{total} = L_{data} + 1.00$ . Why this step? $λ$ says how loudly this rule shouts versus the raw data fit.
Read the gradient direction. In Cell A the whole symbolic penalty is $L_{symbolic} = λ [a - b]_{+} = λ (a - b)$ (because $a > b$ , the floor is inactive). Differentiating that expression: $\partial L_{symbolic} / \partial a = + λ = + 2$ (push $a$ down) and $\partial L_{symbolic} / \partial b = - λ = - 2$ (push $b$ up). The factor $λ = 2$ comes from the weight, not from the subtraction. Why this step? Backprop moves parameters opposite the gradient, so pneumonia-confidence falls and opacity-confidence rises until they agree; naming $λ$ explicitly shows why the slope is $\pm λ$ and not $\pm 1$ .

Verify: units — probabilities are dimensionless, so the penalty is a pure number, addable to $L_{data}$ . Sanity: if the network later reports $p (opacity_{L}) = 0.85 \geq 0.80$ , gap becomes negative, floor fires, penalty $\to 0$ — exactly Cell B/C below.

The figure above shows Ex 1 geometrically: the two sliders $a$ (pneumonia) and $b$ (opacity) on a bar. The blue gap between them is the violation; the arrows show backprop pulling $a$ down and $b$ up until the bars meet and the gap closes.

Worked example Ex 2 — Cells B & C: satisfied and over-satisfied (penalty

$= 0$ , dead gradient) Statement. Same rule as Ex 1. Case (i): $a = 0.40, b = 0.40$ . Case (ii): $a = 0.30, b = 0.90$ . Compute the violation in each and explain what the gradient does.

Forecast: both give penalty $0$ — but are they the same kind of zero?

Case (i) gap. $a - b = 0.40 - 0.40 = 0.00$ ; $[0]_{+} = 0$ . This is Cell B — exactly satisfied, sitting on the boundary. Why this step? The rule is just barely obeyed; we want to see whether training still pushes here.
Case (ii) gap. $a - b = 0.30 - 0.90 = - 0.60$ ; $[- 0.60]_{+} = 0$ . This is Cell C — over-satisfied ( $b$ far exceeds $a$ ). Why this step? The conclusion is more true than the premise demands; logically fine, so no penalty.
Gradient in both. For any point with $a < b$ (Cell C) or $a = b$ from below, the derivative of $[a - b]_{+}$ w.r.t. both $a, b$ is $0$ . Why this step? A zero gradient means the symbolic loss is silent here — the network is only shaped by $L_{data}$ .
Interpret the danger. In Cell C the floor masks a large slack ( $0.60$ ). If your data later drifts and pushes $a$ up past $b$ , the rule can silently break — you only notice when it re-enters Cell A. Why this step? This is the practical reason to log violations, not just the loss.

Verify: plug back. Cell B: $λ \cdot 0 = 0$ added, $L_{total} = L_{data}$ . Cell C: same. Both correct — the two zeros are numerically equal but Cell C hides a $0.60$ buffer.

The figure above is the whole story of Cells A–C: the floor $[a - b]_{+}$ against the gap $a - b$ . Look at the pink flat floor on the left — that is where Cells B and C live (gradient dead). The blue rising ramp on the right is Cell A (gradient $= 1$ before weighting, rule actively corrected). The kink at $0$ is where a rule flips from "shouting" to "silent".

Worked example Ex 3 — Cell D: chained soft-AND, confidence shrinks

Statement. A rule needs three premises all true: $μ_{P} = 0.9$ , $μ_{Q} = 0.8$ , $μ_{R} = 0.7$ . Using soft-AND (product), what is the joint truth degree? Then compare to hard-AND ( $min$ ).

Forecast: will the product be near the smallest input ( $0.7$ ) or noticeably below it?

Multiply the sliders. $0.9 \times 0.8 \times 0.7 = 0.504$ . Why this step? Soft-AND is repeated multiplication; three factors below $1$ compound.
Compare hard-AND. $min (0.9, 0.8, 0.7) = 0.70$ . Why this step? The classical operator only tracks the weakest link; the product punishes every uncertainty.
Read the gap. $0.70 - 0.504 = 0.196$ — the product is $\approx 0.2$ lower. Why this step? This is the "vanishing confidence" trap: chain 10 premises of $0.9$ each and you get $0. 9^{10} \approx 0.349$ , so a long rule looks nearly false even when each part is strong.
Why the product anyway? It is smooth everywhere, so gradients flow to all premises, whereas $min$ only passes gradient to the single smallest term. Why this step? Trainability beats calibration here; practitioners rescale or use temperature to fight the shrink.

Verify: $0.9 \cdot 0.8 = 0.72$ ; $0.72 \cdot 0.7 = 0.504$ ✓. And $0. 9^{10} = 0.34867844 \dots$ ✓ (checked in VERIFY).

The figure above plots how fast $0. 9^{k}$ decays as the chain length $k$ grows: even strong-but-uncertain premises collapse toward $0$ , contrasted with the flat $min$ line that ignores chain length.

Worked example Ex 4 — Cell E: two-sided mutual exclusivity

Statement. A sleep monitor must never call someone both "awake" and "asleep". Network gives $p (awake) = 0.7$ , $p (asleep) = 0.6$ . The rule "exactly one of two exclusive classes" gives violation $[p (c_{1}) + p (c_{2}) - 1]_{+}$ . Compute it, and the corrected minimal-move probabilities.

Forecast: the two probabilities sum to $1.3 > 1$ — how much mass is illegal?

Sum the mass. $0.7 + 0.6 = 1.3$ . Why this step? Mutually exclusive true probabilities can sum to at most $1$ ; the excess is the contradiction.
Violation. $[1.3 - 1]_{+} = 0.30$ . Nonzero — active Cell E. Why this step? This is a two-sided push: gradient $+ 1$ on both classes, so both get pulled down.
State the optimisation criterion, then correct. We seek the least-norm fix: the smallest change $(δ_{1}, δ_{2})$ (minimising $δ_{1}^{2} + δ_{2}^{2}$ ) that restores $p_{1} + p_{2} \leq 1$ . Minimising a sum of squares subject to $δ_{1} + δ_{2} = 0.30$ has its solution at $δ_{1} = δ_{2} = 0.15$ (equal split), so each probability drops $0.15$ → $p (awake) = 0.55$ , $p (asleep) = 0.45$ , sum $= 1.00$ . Why this step? "Split evenly" is not arbitrary — it is exactly what minimising the squared adjustment forces, since the constraint is symmetric in both variables. That is also why the raw gradient (equal $+ 1$ on both) points that way.
Contrast with Ex 1. Implication (Ex 1) is one-sided (down on $a$ , up on $b$ ); exclusivity is two-sided (down on both). Same floor $[\cdot]_{+}$ , opposite geometry. Why this step? Recognising the direction pattern is the exam-relevant skill.

Verify: $0.55 + 0.45 = 1.00$ ✓, violation now $[0]_{+} = 0$ ✓ (Cell B reached). Least-norm check: $0.1 5^{2} + 0.1 5^{2} = 0.045$ is smaller than any unequal split summing to $0.30$ (e.g. $0.3 0^{2} + 0 = 0.09$ ) ✓.

The figure above shows the two stacked probability bars overshooting the $1.0$ ceiling by $0.30$ (pink excess), then the equal $0.15$ trims that bring the stack exactly to the line.

Worked example Ex 5 — Cells F & G: degenerate facts

$μ = 0$ and $μ = 1$ Statement. Rule: $parent (X, Y) \land parent (Y, Z) \Rightarrow grandparent (X, Z)$ . Facts: $μ_{parent(Alice,Bob)} = 1.0$ , $μ_{parent(Bob,Cara)} = 1.0$ , but $μ_{parent(Bob,Dan)} = 0.0$ (unknown/false). Derive $μ_{grandparent(Alice,Cara)}$ and $μ_{grandparent(Alice,Dan)}$ . Use the merge operator $\lor$ (the OR-aggregator defined above) to fold a newly derived degree into the current fact base.

Forecast: one grandparent link should fire fully, the other should stay dead — which is which?

Cara path (Cell G). Soft-AND: $1.0 \times 1.0 = 1.0$ . Why this step? Both premises certain → product certain; the rule fires cleanly.
Merge into the fact base. The prior belief was $μ_{grandparent(Alice,Cara)}^{(0)} = 0$ ; merge with the new evidence: $0 \lor 1.0 = 1.0$ . Why this step? This is the merge $\lor$ (best evidence so far), not the floor $[\cdot]_{+}$ — both arguments are truth degrees, and it promotes the fact from "unknown" to "true".
Dan path (Cell F). Soft-AND: $1.0 \times 0.0 = 0.0$ . Why this step? A single $μ = 0$ premise annihilates the product — "both must hold" fails when one is unknown.
Merge into the fact base. $0 \lor 0.0 = 0.0$ — nothing is inferred. Why this step? The merge keeps the strongest evidence, which is still $0$ ; degenerate zero input correctly produces zero output, no phantom knowledge.

Verify: $1 \cdot 1 = 1$ ✓, $1 \cdot 0 = 0$ ✓. Multiplication by $0$ is the differentiable analogue of "false AND anything = false", exactly matching hard logic. Merge with prior $0$ leaves each result unchanged, as expected.

Worked example Ex 6 — Cell H: forward chaining to a fixed point (limiting behaviour)

Statement. Rule: $parent (X, Y) \Rightarrow ancestor (X, Y)$ and $parent (X, Y) \land ancestor (Y, Z) \Rightarrow ancestor (X, Z)$ . Facts: $parent(A,B) = 1$ , $parent(B,C) = 1$ . At every round we merge ( $\lor$ ) newly derived degrees into the fact base. Iterate until no fact changes; how many rounds?

Forecast: guess the number of iterations before the fact set stops growing.

Round 0. Only the two parent facts, all ancestor degrees $0$ . Why this step? Establish the initial fixed-point candidate.
Round 1. Base rule: $ancestor(A,B) = 0 \lor 1 = 1$ , $ancestor(B,C) = 0 \lor 1 = 1$ . Why this step? Every parent becomes an ancestor first (shortest chains); the merge $\lor$ promotes both from $0$ to $1$ .
Round 2. Transitive rule: $parent(A,B) \land ancestor(B,C) = 1 \cdot 1 = 1$ , so $ancestor(A,C) = 0 \lor 1 = 1$ . Why this step? Now longer chains close; $A$ reaches $C$ .
Round 3. Re-apply all rules: every merge returns the value already stored (e.g. $1 \lor 1 = 1$ ), so no membership degree changes. Fixed point reached. Why this step? Convergence is defined as a round where $F^{(t + 1)} = F^{(t)}$ — you stop when the merge produces no change, giving $3$ rounds (the last one confirms).

Verify: derived facts ${ancestor(A,B), ancestor(B,C), ancestor(A,C)}$ , sizes per round $2, 4, 5, 5$ ; growth halts at round $3$ ✓. The limiting number of ancestor facts for a chain of $3$ people is $(2 3) = 3$ ✓.

Worked example Ex 7 — Cell I: real-world word problem (end-to-end loan pipeline)

Statement. A loan model must satisfy: "IF applicant is flagged high-risk THEN loan must be rejected." Neural scorer gives $p (high-risk) = 0.85$ , $p (approve) = 0.70$ . Encode the rule as a violation, compute it with $λ = 3.0$ , and state the interpretable explanation the XAI layer prints.

Forecast: high-risk and likely-approve — the rule is broken; how big is the penalty?

Rewrite rule. "high-risk $\Rightarrow$ reject" is the same as "high-risk $\Rightarrow$ NOT approve", so $b = \neg p (approve) = 1 - 0.70 = 0.30$ . Why this step? Implication needs a "then" truth degree; the conclusion is the negation of approval, using soft-NOT.
Gap. $a - b = 0.85 - 0.30 = 0.55$ . Why this step? Premise (high-risk) outruns the reject-degree by $0.55$ .
Violation & loss. $[0.55]_{+} = 0.55$ ; $λ \cdot 0.55 = 3.0 \times 0.55 = 1.65$ added to $L_{data}$ . Why this step? Heavy $λ$ because a compliance rule must dominate raw score-fitting.
Explanation string. XAI prints: "Rejected: high-risk=0.85 triggers the reject rule; approval was suppressed by symbolic constraint (violation 0.55)." Why this step? The symbolic side gives a human-readable reason, which "neuron 4738 fired" never could.

Verify: $1 - 0.70 = 0.30$ ✓, $0.85 - 0.30 = 0.55$ ✓, $3.0 \cdot 0.55 = 1.65$ ✓. Once the network lowers $p (approve)$ to $\leq 0.15$ , $b \geq 0.85$ , gap $\leq 0$ , penalty $\to 0$ — compliant.

The figure above traces the loan pipeline left-to-right: raw features → neural scorer → two probabilities → symbolic rule box → floored violation $0.55$ → weighted penalty $1.65$ → human-readable reject reason.

Worked example Ex 8 — Cell J: the exam twist (sign trap + which

$λ$ wins) Statement. Two rules apply at once. Rule 1 (implication): $a_{1} = 0.4, b_{1} = 0.9$ , weight $λ_{1} = 5$ . Rule 2 (exclusivity): $p (c_{1}) = 0.55, p (c_{2}) = 0.60$ , weight $λ_{2} = 1$ . Which rule contributes to the loss, and what is the total symbolic loss? (Trap: don't assume the big $λ$ dominates.)

Forecast: the heavier rule has $λ = 5$ — but does it actually fire?

Rule 1 violation. $[0.4 - 0.9]_{+} = [- 0.5]_{+} = 0$ . Why this step? Conclusion exceeds premise → Cell C → the floor $[\cdot]_{+}$ returns zero, no matter how large $λ_{1}$ is.
Rule 1 contribution. $λ_{1} \cdot 0 = 5 \times 0 = 0$ . Why this step? This is the trap: a huge weight on a satisfied rule contributes nothing.
Rule 2 violation. $[0.55 + 0.60 - 1]_{+} = [0.15]_{+} = 0.15$ . Why this step? Exclusivity mass exceeds $1$ → Cell E fires.
Total. $L_{symbolic} = 0 + λ_{2} \cdot 0.15 = 1 \times 0.15 = 0.15$ . Why this step? Only the small- $λ$ rule matters, because it is the only one violated. Weight scales an active penalty; it cannot resurrect a dead one.

Verify: $0.4 - 0.9 = - 0.5 \Rightarrow 0$ ✓; $0.55 + 0.60 - 1 = 0.15$ ✓; total $= 0.15$ ✓. Lesson: always compute the floor $[\cdot]_{+}$ before multiplying by $λ$ .

Recall Self-test — cover the answers

Ex 1 penalty at $λ = 2$ , gap $0.5$ ? ::: $1.00$ , and $a$ is pushed down / $b$ up; slope is $\pm λ = \pm 2$ . The two zeros in Ex 2 — same value, different meaning? ::: Yes: Cell B sits on the boundary, Cell C hides a $0.60$ slack; both give gradient $0$ . Soft-AND of $0.9, 0.8, 0.7$ ? ::: $0.504$ (vs $min = 0.70$ ). Exclusivity violation for $0.7 + 0.6$ ? ::: $0.30$ , push both down by $0.15$ (least-norm split). The floor $[\cdot]_{+}$ vs the merge $\lor$ — same thing? ::: No: $[\cdot]_{+}$ is a one-sided penalty floor (one input is $0$ ); $\lor$ is the two-input best-evidence OR over truth degrees. Is $p (cat)$ different from $μ_{cat}$ ? ::: No — identical $[0, 1]$ slider; $p$ names a classifier output, $μ$ names a logic truth degree. Grandparent link with a $μ = 0$ premise? ::: $0$ — nothing inferred (Cell F). Forward-chaining rounds for a 3-person chain? ::: $3$ (last round confirms fixed point). Loan penalty at $λ = 3$ ? ::: $1.65$ . Ex 8 total symbolic loss? ::: $0.15$ — the big- $λ$ rule was satisfied, so contributed $0$ .

Related deep dives to visit next: 4.3.05-Attention-Mechanisms (how NMN modules route attention), 6.5.07-Reinforcement-Learning-Advanced (symbolic reward shaping), 6.5.10-Causal-Inference (rules as interventions). Back to the parent topic.