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Proof step 1 from proof null-not-start-of-path (conclusion)

⊦ ¬ path(null ' + x, 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 [], +;;%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%app: ⊦ x0 + [] = x0; from specification list[reflist]