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For the symmetric group on five letters, $S^{(4,1)}$ is the four-dimensional augmentation representation. In characteristic three it is simple, since its only possible proper nonzero submodule would be the constant line, which is excluded by $5\ne 0$ in the field. Thus it is $D^{(4,1)}$. If this module occurs as a submodule of $S^{(3,2)}$, the quotient has dimension one, since the standard polytabloid basis gives $\dim S^{(3,2)}=5$.

A transposition has trace one on $S^{(3,2)}$, by the Murnaghan–Nakayama rule, and trace two on $S^{(4,1)}$, by subtracting the trivial line from the five-point permutation representation. These integral traces reduce to the same elements in characteristic three. The quotient trace is therefore minus one, so its one-dimensional action is the sign representation. In particular an arbitrary permutation acts on it by its sign. This identifies the quotient action without asserting that the original short exact sequence splits.

A first-order sentence consisting of a block of universal quantifiers followed by a quantifier-free formula, obtained by Skolemization in an expanded first-order signature. Its matrix can additionally be put into conjunctive normal form to obtain clausal form.

If a first-order theory proves both $A$ and $\neg A$, the external truth of its Rosser provability predicate for $A$ depends on which side has the earlier formal proof code. A reserved least code for a formal proof of $\neg A$ makes the Rosser assertion false. Reserving the least code for a formal proof of $A$ makes it true. These are legitimate effective renumberings of the same formal proofs. Inconsistent first-order theories therefore require an explicit proof-order convention.

Let $M=4$ be a Finite von Neumann ordinal and $N=V_4$. With parameters $x=3$, $y=2$, the transitive structure $M$ satisfies the first-order power-set specification, since its available subsets of $2$ are exactly $0,1,2$. In $N$, the additional subset $\{1\}$ makes the same specification false. Bounded logical formulas are absolute between transitive sets by induction, so existential logical formulas with bounded matrices are upward absolute. This counterexample therefore rules out an unrestricted logical equivalence of power-set equality with a ${\Sigma }_1$ logical formula.

For bounded positive operators satisfying the hypotheses, $(S-T)(S+T)=0$, so $S-T$ vanishes on the closure of the range of $S+T$. The kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Self-adjointness and the Hilbert projection theorem decompose the whole Hilbert space as that closed range plus that kernel. Thus $S-T=0$. This proves the commuting case of uniqueness of a positive square root without using a spectral representation.