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Proof step 11 from proof in-mod (previous step)

sdt0[str][p0] = dnode(md0, sdt), q = str0 + q0, p0 ∈ sdt0[str], str ∈ sdt0, ∀ md, sdt0, md0, sdt, q, md1. dnode(md, sdt0)[p0] = dnode(md0, sdt) ∧ p0 ∈ dnode(md, sdt0) → (q ∈ dnode(md, sdt0)[p0, dnode(md1, sdt)] ↔ q ∈ dnode(md, sdt0)) ∨ p0 = ε ⊦ q ∈ dnode(md, sdt0)[str ' + p0, dnode(md1, sdt)] ↔ q ∈ dnode(md, sdt0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]
Proof step 13:
⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str]
Proof step 14:
sdt0[str][p0] = dnode(md0, sdt), str ∈ sdt0, p0 ∈ sdt0[str], ∀ md, sdt0, md0, sdt, q, md1. dnode(md, sdt0)[p0] = dnode(md0, sdt) ∧ p0 ∈ dnode(md, sdt0) → (q ∈ dnode(md, sdt0)[p0, dnode(md1, sdt)] ↔ q ∈ dnode(md, sdt0)) ∨ p0 = ε ⊦ str0 ' ∈ dnode(md, sdt0)[str ' + p0, dnode(md1, sdt)] ∧ q0 ∈ dnode(md, sdt0)[str ' + p0, dnode(md1, sdt)][str0 '] ↔ str0 ∈ sdt0 ∧ q0 ∈ sdt0[str0]

Rule argument

Additional information

Used simplifier rules: split-in: ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p];%in-dnode: ⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str];%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str + p = str ' + p; (s)