Turing machines formulas

Only accepted strings count; strings on which M loops or rejects are outside L(M).

Here the complement is taken with respect to ${\Sigma }^{\ast }$. To decide, run both recognisers in parallel (alternating steps); exactly one of them accepts: the one for $L$ if $w\in L$, the one for $\overline{L}$ otherwise.

Use the contrapositive: if $A{\le }_{m}B$ and $A$ is undecidable, $B$ is undecidable. Also: $A{\le }_{m}B$ and $B$ RE implies $A$ RE; $A{\le }_{m}B$ and $A$ not RE implies $B$ not RE. Since $A{\le }_{m}B$ if and only if $\overline{A}{\le }_m\overline{B}$, reducing $A_{TM}$ to $\overline{B}$ proves $B$ is not RE. Direction matters.

P is a property of the language L(M). Non-trivial: some TM languages have it and some do not. Semantic: it depends only on L(M), not on the machine's structure.