Where in C++14 Standard does it say that a non-constexpr function cannot be used in a definition of a constexpr function?

Question

For example the code below doesn't compile unless incr() is declared constexpr:

int incr(int& n) {
    return ++n;
}

constexpr int foo() {
    int n = 0;
    incr(n);
    return n;
}

Looking at §7.1.5/3 in C++14 we have:

The definition of a constexpr function shall satisfy the following constraints:
(3.1) — it shall not be virtual (10.3);
(3.2) — its return type shall be a literal type;
(3.3) — each of its parameter types shall be a literal type;
(3.4) — its function-body shall be = delete, = default, or a compound-statement that does not contain

(3.4.1) — an asm-definition,
(3.4.2) — a goto statement,
(3.4.3) — a try-block, or
(3.4.4) — a definition of a variable of non-literal type or of static or thread storage duration or for which no initialization is performed.

Answer 1

Two paragraphs later, in [dcl.constexpr]/5:

For a non-template, non-defaulted constexpr function or a non-template, non-defaulted, non-inheriting constexpr constructor, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression (5.20), or, for a constructor, a constant initializer for some object (3.6.2), the program is ill-formed; no diagnostic required.

No argument exists such that foo() could be a core constant expression because of incr(), therefore the program is ill-formed (NDR).