$NP\subseteq PSPACE\subseteq EXP$

To prove $NP\subseteq PSPACE$, let $L\in NP$. By [L1], choose a verifier for $L$ with runtime bounded by a polynomial $q(n)$ and certificate length bounded by a polynomial $p(n)$. A deterministic decider for $L$ can enumerate all certificates $u$ of length at most $p(\mathrm{\mid }x\mathrm{\mid })$ in lexicographic order, run the verifier on $(x,u)$, and accept as soon as one certificate succeeds. It stores only the input $x$, the current certificate $u$, and the verifier workspace, so the total space used is polynomial in $\mathrm{\mid }x\mathrm{\mid }$. Hence $L\in PSPACE$ by [L2] and [L3].

To prove $PSPACE\subseteq EXP$, let $L\in PSPACE$. Then by definition there exists $k\ge 1$ and a deterministic machine $M$ deciding $L$ while using at most $c{n}^k$ tape cells on inputs of length $n$, for some constant $c$. Fix such an input length $n$. Because of [L5], a configuration of $M$ is determined by one of finitely many states, the head positions on the finitely many tapes, and the contents of at most $c{n}^k$ visited cells. Therefore the total number of possible configurations is at most $\mathrm{\mid }Q\mathrm{\mid }\cdot (c{n}^k+1{)}^d\cdot \mathrm{\mid }\mathrm{\Gamma }{\mathrm{\mid }}^{dc{n}^k}=2^{O(n^k)}$ for a constant $d$ depending only on the fixed number of tapes.

Since $M$ is deterministic and halts on every input, it cannot visit the same configuration twice on one accepting or rejecting run: a repeated configuration would force an infinite loop from that point onward. So the running time on inputs of length $n$ is bounded by the number of reachable configurations from step 1.2, hence by $2^{O(n^k)}$. Enlarging the exponent if necessary, there exists $m$ with $2^{O(n^k)}\le 2^{n^m}$ for all sufficiently large $n$. Therefore $L\in EXP$ by [L2] and [L4].

Steps 1.1 and 2.1 prove $NP\subseteq PSPACE\subseteq EXP$.