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Proof step 3 from proof path-first-in-H (previous step)

path(r ' + x, H) ⊦ r ∈ H

User applied rule constructor cut

And closed the goal!

Rule argument

var: x

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ r ∈ H ∧ r ≠ null;%path-cons: ⊦ path(r ' + r0 ' + x, H) ↔ r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(r0 ' + x, H);%reflist freely generated by [], +;;% ⊦ r + x0 = r ' + x0; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ x0 + [] = x0; (s)