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Proof step 46 from proof at-mod (previous step)

sdt[str] = dnode(md0, sdt0), q0 ∈ sdt[str0], str ∈ sdt, str0 ∈ sdt, ∀ 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]) ⊦ sdt[str0][q0] = (str0 ' + q0 = str ' ⊃ sdt[str0][q0][restn(# q0, ε), dnode(md1, sdt1)];sdt[str0][q0]), str0 = str

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ str ' = str0 ' + p ↔ str = str0 ∧ p = ε; (ws)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]