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Proof step 6 from proof at-explicit-other (previous step)

st[m] = mkblock(x, lax, m2), m ∈ st, @(m0, m1) ∈ lax ⊦ st[n, ablock][@(m, m0, m1)] = x[pos(@(m0, m1), lax)], m = n, @(m, m0, m1) ∈ st[n, ablock]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information