Two empty work tapes can be merged into one

A predicate decidable in polynomial time with $s+2$ blank work tapes is decidable in polynomial time with $s+1$.

This is the easy half of tape elimination, and it is easy for one specific reason: both tapes being merged start empty, so there is nothing to gather and no existing content to re-encode. The two may simply be hosted as two tracks of a single tape, each position carrying one cell of each track together with a flag recording whether that track's head is there. Since exactly one physical head must serve two virtual heads, each simulated step costs a sweep of the region written so far; that region grows by at most one position per step, so the overhead is quadratic — comfortably polynomial.

The statement should not be mistaken for a warm-up exercise. Avoiding the re-encoding phase does not avoid the per-step simulation argument: the flags, the sweeps, and the invariant relating the simulated configuration to the physical one are all still required. It is the full construction minus one phase.