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Proof step 6 from proof put-fnode (previous step)

dnode(md, sdt)[p, tree0] = fnode(fid), p ∈ dnode(md, sdt) ⊦ p = ε ∧ tree0 = fnode(fid)

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
⊦ tree[ε, tree0] = tree0
Proof step 8:
⊦ ε ∈ tree
Proof step 9:
dnode(md, sdt)[p, tree0] = fnode(fid), p = str + p0, p ∈ dnode(md, sdt) ⊦ p = ε ∧ tree0 = fnode(fid)

Rule argument

var: p

Additional information

Used simplifier rules: nil-put: ⊦ tree[ε, tree0] = tree0;%nil-in: ⊦ ε ∈ tree;%path freely generated by ε, +;;