Let \(z\) be a symbol and \(c\) a configuration of a \(k\)-tape machine on an input word \(x\). The symbol scanned by the input head of the masked configuration on \(z^{\left\lvert x\right\rvert }\) is blank exactly when the symbol scanned in \(c\) is blank, and equals \(z\) whenever \(c\) scans any nonblank symbol.

Let \(z\) be a symbol, \(a\) an action (an input-head move, a write and a move for each work tape, an optional emitted symbol, and a next state), and \(c\) a configuration on an input word \(x\). Then masking the input by \(z\) after applying \(a\) yields the same configuration as applying \(a\) to the masked configuration.

Let \(W\) be a machine over an alphabet \(A\), let \(z \in A\), and let \(c\) be a configuration of the masked clock of \(W\) on an input word \(x\). Masking the result of one step of the masked clock at \(c\) yields the same configuration as performing one step of \(W\) at the masked configuration of \(c\); this includes the absorbing case in which \(c\) is already halted.

Let \(W\) be a machine over an alphabet \(A\), let \(z \in A\), let \(x\) be an input word, and let \(t \in \mathbb {N}\). Masking the configuration reached by the masked clock of \(W\) after \(t\) steps from its initial configuration on \(x\) yields exactly the configuration reached by \(W\) after \(t\) steps from its initial configuration on the constant input \(z^{\left\lvert x\right\rvert }\) — an equality of all five configuration components (state, input position, work tapes, work-head positions, and output), at live and halted times alike.

For every machine \(W\) over an alphabet \(A\) and every symbol \(z \in A\), the masked clock of \(W\) with mask \(z\) is oblivious: on any two inputs of the same length its input-head position and all of its work-head positions agree at every time. No halting hypothesis is required.

Let \(P\) be a machine over an alphabet \(A\) and let \(z \in A\). Suppose the transition table of \(P\) is insensitive to nonblank input contents: for every state \(q\), scanned input symbol \(s\), and tuple of work-tape reads \(w\), the transition of \(P\) on the read obtained from \(s\) by replacing any nonblank symbol with \(z\) coincides with the transition of \(P\) on \(s\) itself. Then \(P\) is oblivious.

Let \(W\) be a machine over the Boolean alphabet, \(T : \mathbb {N} \to \mathbb {N}\), and \(b \in \mathbb {N}\). Suppose that on every input \(x\), after \(b \cdot (T(\left\lvert x\right\rvert ) + 1)\) steps \(W\) has halted with output exactly the list of binary digits of \(T(\left\lvert x\right\rvert )\). Then the masked clock of \(W\) (with mask the symbol \(\mathsf{false}\)) satisfies the same specification: on every input \(x\) it halts within \(b \cdot (T(\left\lvert x\right\rvert ) + 1)\) steps with output the binary digits of \(T(\left\lvert x\right\rvert )\).

Let \(P\) be a machine over an alphabet \(A\) whose transition table never emits an output symbol, and consider its decoration by any data transducer with \(l\) extra tapes, designated head, initial data state, and visit function. For every configuration \(c\) of the decorated machine, the schedule projection of one step of the decorated machine at \(c\) equals one step of \(P\) at the schedule projection of \(c\). This is an equality of entire configurations, not merely a correspondence at macrostep boundaries.

Let \(P\) be a machine over an alphabet \(A\) whose transition table never emits an output symbol, decorated by any data transducer. For every input word \(x\) and every time \(t \in \mathbb {N}\), the schedule projection of the decorated machine’s configuration after \(t\) steps from its initial configuration on \(x\) equals the configuration of \(P\) after \(t\) steps from its initial configuration on \(x\).

Let a machine \(P\) over an alphabet \(A\) be decorated by any data transducer with \(l\) extra tapes and designated head \(h\). For every input word \(x\), every time \(t \in \mathbb {N}\), and every extra-tape index \(i \lt l\), the head position of the \(i\)-th data tape of the decorated machine after \(t\) steps equals the head position of the designated schedule head \(h\) at the same time — including at all times after halting. Consequently no data-selected write can alter a head trajectory.

Let \(P\) be a machine over an alphabet \(A\) whose transition table never emits an output symbol, and suppose \(P\) is oblivious. Then for every choice of data transducer — number of extra tapes, designated head, finite data-state set with initial element, and visit function — the decorated machine is oblivious. The data states, emitted answers, and data-selected writes may depend arbitrarily on the actual input; none of them can choose a head movement or terminate the schedule early.

For every possibly-blank symbol \(a\) over an alphabet \(A\) with decidable equality, decoding the transverse encoding of \(a\) returns \(a\); this includes the blank block.

For every embedding \(e\) of the Booleans into an alphabet \(A\) with decidable equality and every bit \(b\), decoding the embedded symbol \(e(b)\) recovers \(b\).

Let \(c\) be a configuration of a machine over a finite alphabet \(A\) on the symbolwise embedding of a binary word \(x\) under an embedding \(e\) of the Booleans into \(A\). Mapping the (possibly blank) input bit scanned by the encoded configuration through \(e\) recovers exactly the (possibly blank) input symbol scanned by \(c\).

Let \(c\) be a configuration of a \(k\)-tape machine over a finite alphabet \(A\) on the symbolwise embedding of a binary word. For every logical tape index \(i \lt k\), decoding the family of bits scanned on the tracks of block \(i\) in the encoded configuration yields exactly the symbol scanned by \(c\) on tape \(i\).

Let \(P\) be a machine over a finite alphabet \(A\), \(e\) an embedding of the Booleans into \(A\), and \(c\) a configuration of \(P\) on the symbolwise embedding of a binary word \(x\). Then one step of the transverse binary realization of \(P\) at the encoded configuration of \(c\) equals the encoded configuration of one step of \(P\) at \(c\): both the read and the write of a logical cell happen simultaneously across its fixed collection of tracks, and the halted case is absorbing on both sides.

Let \(P\) be a machine over a finite alphabet \(A\), \(e\) an embedding of the Booleans into \(A\), and \(x\) a binary word. Encoding the initial configuration of \(P\) on the symbolwise embedding of \(x\) yields exactly the initial configuration of the transverse binary realization of \(P\) on \(x\).

Let \(P\) be a machine over a finite alphabet \(A\), \(e\) an embedding of the Booleans into \(A\), \(x\) a binary word, and \(t \in \mathbb {N}\). The configuration of the transverse binary realization of \(P\) after \(t\) steps from its initial configuration on \(x\) equals the encoding of the configuration of \(P\) after \(t\) steps from its initial configuration on the symbolwise embedding of \(x\) — with no intermediate coding states and no variable-duration memory accesses.

Let \(P\) be a machine over a finite alphabet \(A\) and \(e\) an embedding of the Booleans into \(A\). If \(P\) is oblivious, then the transverse binary realization of \(P\) is oblivious — at all intermediate physical times and at all times after halting.

Let \(P\) be a machine over a finite alphabet \(A\), \(e\) an embedding of the Booleans into \(A\), \(x\) and \(w\) binary words, and \(t \in \mathbb {N}\). If, started on the symbolwise embedding of \(x\), the machine \(P\) has halted after \(t\) steps with output exactly the symbolwise embedding of \(w\), then the transverse binary realization of \(P\), started on \(x\), has halted after the same \(t\) steps with output exactly \(w\).

For every clock witness \(W\) over the Boolean alphabet, every parameter \(a \in \mathbb {N}\), every state \(q\) of the length-only schedule, every scanned input symbol, and every tuple of work-tape reads, the transition of the length-only schedule emits no output symbol. A decoration supplies the final answer at the schedule’s final halting transition.

For every clock witness \(W\) over the Boolean alphabet and every parameter \(a \in \mathbb {N}\), the length-only schedule built from \(W\) and \(a\) is oblivious: on arbitrary equal-length inputs over the schedule alphabet, its input head and every clock, budget, counter, and guide head occupy the same position at every time. No computation or halting hypothesis is needed.