Under diamond, ccc fails to survive a square

F1 supplies the tree under the stated diamond and A1 hypotheses. Its successor sets contain at least two ordinal-coded nodes, so their first and second members define $t^0,t^1$ without further choices. The split-pair construction in F2 applies to exactly these selections. Each first successor determines its parent, so $t\mapsto (t^0,t^1)$ is injective and $\mathrm{\mid }E\mathrm{\mid }={\mathrm{\aleph }}_1$. For incomparable parents even the first successors are incompatible; for $t{<}_{T}u$, simultaneous coordinate compatibility would put the two distinct $t$-successors below $u$, impossible by unique predecessors, as calculated in F2. Thus the displayed $E$ is the explicit antichain, and $P$ is ccc.

If $P$ were Knaster, F3 would make $P\times P$ Knaster and F4 would make it ccc. This contradicts the antichain $E$ in step 1.1. Hence this conditional ccc example is not Knaster; the diamond assumption remains necessary for the tree supplied here.