The graded shuffle coproduct is coassociative, cocommutative, and conilpotent

The quotient relations are generated by adjacent graded swaps. In each unshuffle summand of [F1], swapping adjacent inputs either changes their order inside one output block, where the same quotient relation applies, or moves them between blocks, where the graded flip in the target tensor product contributes exactly $(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}$. The unshuffle terms are reindexed by this swap, so [F1] and [F2] give the same output class; hence $\bar{\mathrm{\Delta }}$ descends to $S^n(W)$.

Both iterated coproducts $(\bar{\mathrm{\Delta }}\otimes 1)\bar{\mathrm{\Delta }}$ and $(1\otimes \bar{\mathrm{\Delta }})\bar{\mathrm{\Delta }}$ on a length-$n$ tensor sum over the same ordered decompositions of its labels into three nonempty, internally ordered blocks. In either parenthesization the sign is the Koszul sign of the single permutation that places those labels into the same three blocks. The terms therefore agree one by one, proving coassociativity.

The graded flip interchanging the two tensor factors sends a $(p,n-p)$-unshuffle to the complementary $(n-p,p)$-unshuffle. Its block-swap sign is exactly the Koszul sign for moving every factor in the first block past every factor in the second. Thus the terms pair with identical coefficients after the graded flip, proving graded cocommutativity.

Each application of $\bar{\mathrm{\Delta }}$ splits a length-$n$ tensor into two positive lengths. An iterated coproduct into $r$ factors is consequently a sum over ordered decompositions into $r$ nonempty blocks; if $r>n$ there are no such decompositions. Every element of $S^c(W)$ is a finite sum of lengths, so a sufficiently long iterated reduced coproduct kills it.

The length-one component of $\bar{\mathrm{\Delta }}$ on $S^n(W)$ is $S^n(W)\to W\otimes S^{n-1}(W)$. Compose it with concatenation $W\otimes S^{n-1}(W)\to S^n(W)$. For a pure tensor, each of the $n$ choices of the singleton factor returns the same graded-symmetric class: the Koszul sign used to move that factor to the front is canceled by the same sign in concatenating it back. The composite is therefore $n{\mathrm{i}\mathrm{d}}_{S^n(W)}$.

If $x\in S^n(W)$ with $n\ge 2$ is primitive, its reduced coproduct is zero, so the composite in step 1.5 gives $nx=0$. By [F3], multiplication by $n$ is invertible, hence $x=0$. Lengths are a direct sum and every element has finite length support; the only primitives are therefore the length-one elements $W$. Conversely, [F1] gives $\bar{\mathrm{\Delta }}(w)=0$ for each $w\in W$.