Task-aware BDD and SAT advice

This overview explains how advice affects exact decision-diagram and satisfiability lifecycles with checked results and fallback. A smaller diagram or a changed search order does not by itself establish a faster cold or resident request. A decision diagram for and can test A first. A zero immediately determines output zero. When A is one, the answer equals B. The structure encodes both possibilities rather than one fixed assignment. Input zero one reaches zero without needing B to determine the answer. Input one one follows the A-one path and then the B-one path to one. The shorter path for one assignment does not imply that B is irrelevant to the whole function. For a multiplexer, S selects whether the result equals A or B. Testing S first exposes that choice immediately. Other variable orders organize the remaining cases differently and can change the size of a reduced diagram. The useful order is a structural question that needs exact preservation and cost checks. E1 and E2 put advice in front of exact BDD-order and SAT or equivalence lifecycles. An ordering may change diagram size without changing the function. SAT advice may change search without changing what counts as a valid witness. For a conceptual equivalence test, compare A AND B with NOT of NOT A OR NOT B. Build a miter that exclusive-ors their outputs and ask whether it can be true. Exhaustive enumeration finds no satisfying assignment, establishing equivalence on this finite domain. Change the second rule to A OR B and the miter has a witness at 01. A fresh BDD build, a resident restriction session, a fresh SAT solve and an assumptions session have different costs. The retained learned advice did not establish a general timing win and remains unpromoted. Exact witness checking and advice-off fallback stay authoritative. Work a second complete case with A 1, B 0. Keep this assignment fixed throughout the calculation. Each symbol retains the meaning already introduced; a new value does not create a new logical rule. And returns zero because B is false. Or returns one because A is true. Their exclusive-or is one, so this assignment satisfies the miter and demonstrates disagreement. The witness is useful because it can be substituted back into both original expressions and checked directly. The two original formulas are not both true on this assignment: and is zero and or is one. Their miter is true because those truth values differ. Satisfiable means that some satisfying assignment exists. Both and and or are individually satisfiable, for example at one one; that does not make their functions equivalent. The calculated answer belongs to this complete assignment under the stated rule. To compare F and its De Morgan expression, construct their exclusive-or, called a miter. H is one exactly where the two original expressions disagree. Asking whether H can be one is therefore an exact equivalence question. The two output columns agree in all four rows, so the miter is always zero. A sound satisfiability procedure reporting unsatisfiable for the miter establishes that no disagreement assignment exists in the declared universe. Replace G with or and the input zero one makes the miter one. Substitute that witness into the original expressions: and returns zero while or returns one. A satisfying miter assignment explains the actual disagreement. Reusing resident diagrams or solver state has separate preparation and per-query semantics that must be declared. Measure the caller's actual lifecycle and objective. Keep exact witness checks and advice-off fallback authoritative.