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Proof step 5 from proof at-mod (previous step)

dnode(md, sdt)[ε] = dnode(md0, sdt0), q ∈ dnode(md, sdt)[ε, dnode(md1, sdt1)], ε ∈ dnode(md, sdt) ⊦ dnode(md, sdt)[ε, dnode(md1, sdt1)][q] = (q ⊑ ε ⊃ dnode(md, sdt)[q][restn(# q, ε), dnode(md1, sdt1)];dnode(md, sdt)[ε, dnode(md0, sdt1)][q])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ tree[ε] = tree
Proof step 7:
⊦ tree[ε, tree0] = tree0
Proof step 8:
⊦ ε ∈ tree
Proof step 9:
q ∈ dnode(md1, sdt1) ⊦ dnode(md1, sdt1)[q] = (q = ε ⊃ dnode(md1, sdt1);dnode(md0, sdt1)[q])

Rule argument

Additional information

Used simplifier rules: nil-at: ⊦ tree[ε] = tree;%nil-put: ⊦ tree[ε, tree0] = tree0;%nil-in: ⊦ ε ∈ tree;% ⊦ restn(0, p) = p; (s)% ⊦ # ε = 0; (s)% ⊦ p ⊑ ε ↔ p = ε; (s)% ⊦ dnode(md, sdt) = dnode(md0, sdt0) ↔ md = md0 ∧ sdt = sdt0; (ws)