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Proof step 26 from proof in-put (previous step)

sdt0[str0][p0] = dnode(md, sdt), p0 ∈ sdt0[str0], str0 ∈ sdt0, ∀ md0, sdt0, md, sdt, q, str, tree0. dnode(md0, sdt0)[p0] = dnode(md, sdt) ∧ p0 ∈ dnode(md0, sdt0) → (q ∈ dnode(md0, sdt0)[p0, dnode(md, sdt[str, tree0])] ↔ (p0 + str ' ⊑ q ⊃ restn(# p0, q.tail) ∈ tree0;q ∈ dnode(md0, sdt0))) ⊦ str1 ' ∈ dnode(md0, sdt0)[str0 ' + p0, dnode(md, sdt[str, tree0])] ∧ q0 ∈ dnode(md0, sdt0)[str0 ' + p0, dnode(md, sdt[str, tree0])][str1 '] ↔ ((str0 ' + p0) + str ' ⊑ str1 ' + q0 ⊃ restn(# p0, q0.tail) ∈ tree0;str1 ∈ sdt0 ∧ q0 ∈ sdt0[str1]), p0 ≠ ε ∧ str0 ∈ sdt0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
⊦ tree[ε] = tree
Proof step 28:
⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0])
Proof step 29:
⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt
Proof step 30:
⊦ ε ∈ tree
Proof step 31:
⊦ tree[ε, tree0] = tree0
Proof step 32:
sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) → (q ∈ dnode(md2, sdt2[str2, tree]) ↔ (str2 ' ⊑ q ⊃ q.tail ∈ tree;q ∈ dnode(md1, sdt1))) ⊦ str1 ∈ sdt0[str0, dnode(md, sdt[str, tree0])] ∧ q0 ∈ dnode(md0, sdt0[str0, dnode(md, sdt[str, tree0])])[str1 '] ↔ (str0 ' + str ' ⊑ str1 ' + q0 ⊃ q0.tail ∈ tree0;str1 ∈ sdt0 ∧ q0 ∈ sdt0[str1])

Rule argument

Additional information

Used simplifier rules: nil-at: ⊦ tree[ε] = tree;%put-here: ⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0]);%one-ind: ⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt;%nil-in: ⊦ ε ∈ tree;%nil-put: ⊦ tree[ε, tree0] = tree0;% ⊦ restn(0, p) = p; (s)% ⊦ # ε = 0; (s)% ⊦ ε + p = p; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ p + ε = p; (s)