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Proof step 7 from proof path-delnonmember (previous step)

path(r0 ' + x0, H), path(x0, H) → path(x0, H -- r) ⊦ r = r0, path(r0 ' + x0, H -- r), r ∈ x0

User applied rule constructor cut

And closed the goal!

Rule argument

var: x0

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]% ⊦ r ∈ r0 ' + x0 ↔ ¬(r ≠ r0 ∧ ¬ r ∈ x0); (ss)% ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H; (ws)% ⊦ r ≠ r0 → (H -- r0)[r] = H[r]; (s)% ⊦ x0 + [] = x0; (s)% ⊦ ¬ r ∈ []; (s)