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Proof step 45 from proof at-mod (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
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

Rule argument

Additional information

Used simplifier rules: ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)% ⊦ str ≠ str0 → sdt[str0, tree][str] = sdt[str]; (s)