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Proof step 6 from proof path-trans (previous step)

x0 = r0 + x, path(r ' + x0, H), path((r ' + x0).last ' + y, H), ∀ y, H. path(x0.last ' + y, H) ∧ path(x0, H) → path(x0 + y, H) ⊦ path((r ' + x0) + y, H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
path(H[r].nxt ' + x, H), path((H[r].nxt ' + x).last ' + y, H), r ∈ H, ∀ y0, H0. path((H[r].nxt ' + x).last ' + y0, H0) ∧ path(H[r].nxt ' + x, H0) → path((H[r].nxt ' + x) + y0, H0) ⊦ r = null, path(H[r].nxt ' + x + y, H)

Rule argument

Additional information

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