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Proof step 65 from proof at-mod (previous step)

sdt[str][p0] = dnode(md0, sdt0), str0 ∈ sdt[str, sdt[str][p0, dnode(md1, sdt1)]], str ∈ sdt, p0 ∈ sdt[str], q0 ∈ sdt[str0], ∀ md, sdt, md0, sdt0, q, md1, sdt1. dnode(md, sdt)[p0] = dnode(md0, sdt0) ∧ p0 ∈ dnode(md, sdt) ∧ q ∈ dnode(md, sdt)[p0, dnode(md1, sdt1)] → dnode(md, sdt)[p0, dnode(md1, sdt1)][q] = (q ⊑ p0 ⊃ dnode(md, sdt)[q][restn(# q, p0), dnode(md1, sdt1)];dnode(md, sdt)[p0, dnode(md0, sdt1)][q]) ⊦ dnode(md, sdt[str, sdt[str][p0, dnode(md1, sdt1)]])[str0 '][q0] = (str0 ' + q0 ⊑ str ' + p0 ⊃ dnode(md, sdt)[str0 ' + q0][restn(# q0, p0), dnode(md1, sdt1)] ;dnode(md, sdt[str, sdt[str][p0, dnode(md0, sdt1)]])[str0 ' + q0] ), p0 = ε, str0 = str

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: one-at: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str];%at-dnode: ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p];% ⊦ str ' + p ⊑ str0 ' + p0 ↔ str = str0 ∧ p ⊑ p0; (ws)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ≠ str0 → sdt[str0, tree][str] = sdt[str]; (s)% ⊦ str ≠ str0 → (str ∈ sdt[str0, tree] ↔ str ∈ sdt); (s)% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)