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Proof step 39 from proof at-mod (previous step)

sdt[str] = dnode(md0, sdt0), str ∈ sdt, q0 ∈ dnode(md1, sdt1), ∀ md2, sdt2, md3, sdt3, q, md4, sdt4. dnode(md2, sdt2) = dnode(md3, sdt3) ∧ q ∈ dnode(md4, sdt4) → dnode(md4, sdt4)[q] = (q = ε ⊃ dnode(md4, sdt4);dnode(md3, sdt4)[q]) ⊦ dnode(md1, sdt1)[q0] = (q0 = ε ⊃ dnode(md1, sdt1);dnode(md0, sdt1)[q0])

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: md0, sdt0, md0, sdt0, q0, md1, sdt1

Additional information

Used terms: md0, sdt0, md0, sdt0, q0, md1, sdt1