Proof of A First-Order Equation Operator is Degenerate Elliptic and Its $\delta$-Shifts Ignore the Form Argument

Each result cited is universally quantified over the data in its own statement.

Claim 1. Let $x\in W$, let $r\in\mathbb{R}$, let $p\in H$ and let $X,Y\in\mathrm{Sym}(V)$ satisfy $X\preceq Y$. Since $F$ is first order,

$$F(x,r,p,Y)=F(x,r,p,X),$$

and therefore $F(x,r,p,Y)\le F(x,r,p,X)$. As $x$, $r$, $p$, $X$ and $Y$ were arbitrary, $F$ is degenerate elliptic in the sense of Degenerate Elliptic Second-Order Equation Operator on a Hilbert Triple (i).

Claim 2. Let $x\in W$, $r\in\mathbb{R}$, $p\in H$ and $Y,Y'\in\mathrm{Sym}(H)$. By Second-Order Equation Operator on an Open Subset of a Hilbert Triple and Its $\delta$-Shifts (ii) the forms $Y|_{V}+\delta I_{V}$ and $Y'|_{V}+\delta I_{V}$ belong to $\mathrm{Sym}(V)$, and

$$F^{-}_{\delta}(x,r,p,Y)=F\bigl(x,\,r+\delta h(x),\,p+\delta Ax,\,Y|_{V}+\delta I_{V}\bigr),\qquad F^{-}_{\delta}(x,r,p,Y')=F\bigl(x,\,r+\delta h(x),\,p+\delta Ax,\,Y'|_{V}+\delta I_{V}\bigr).$$

The two right-hand sides are values of $F$ at the same first three arguments, hence are equal because $F$ is first order. The same argument applies to $F^{+}_{\delta}$, whose two values are those of $F$ at $\bigl(x,\,r-\delta h(x),\,p-\delta Ax,\,Y|_{V}-\delta I_{V}\bigr)$ and at $\bigl(x,\,r-\delta h(x),\,p-\delta Ax,\,Y'|_{V}-\delta I_{V}\bigr)$, which again differ only in the form argument.