Rice's theorem

WLOG assume the everywhere-undefined function $\emptyset$ (computed by a program that never halts on any input) does not have property $P$ — otherwise argue with the complement property $\lnot P$, which is decidable exactly when $P$ is. Since $P$ is nontrivial, fix some program $e_0$ whose function $\varphi_{e_0}$ does have property $P$.

Suppose, for contradiction, that $P$ is decidable by some algorithm $D$ (given a program index, $D$ halts and correctly reports whether that program's function has property $P$). We reduce the Halting Problem to $P$, contradicting Theorem 1.

Given any pair $(e,x)$, effectively construct (by simple textual manipulation — Kleene's s-m-n theorem) a new program $e'$ that, on any input $y$: first simulates program $e$ running on input $x$; if that simulation ever halts, then $e'$ goes on to simulate program $e_0$ on input $y$ and outputs whatever it outputs.

Now examine the two possibilities. If $e$ halts on $x$: the simulation of $e$ on $x$ finishes, so $e'$ then behaves exactly like $e_0$ on every input, i.e. $\varphi_{e'}=\varphi_{e_0}$ — which has property $P$ (since $P$ is semantic, it only depends on the function computed, and $\varphi_{e_0}$ has $P$). If $e$ does not halt on $x$: the simulation of $e$ on $x$ never finishes, so $e'$ never reaches the $e_0$-simulation step on any input $y$; hence $\varphi_{e'}$ is the everywhere-undefined function $\emptyset$, which by assumption does not have property $P$.

So: $e$ halts on $x$ $\iff$ $\varphi_{e'}$ has property $P$ $\iff$ $D(e')$ answers "yes". Since $(e,x)\mapsto e'$ is computable, the algorithm "compute $e'$ from $(e,x)$, then run $D(e')$" would decide the Halting Problem — contradicting Theorem 1. So no such $D$ exists: $P$ is undecidable. $\blacksquare$