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Proof step 74 from proof rem-stays-reachable-aux2 (previous step)

x2 = x1 + r1 ', x2 = (x + H[r1].nxt ') + r0 ' + y, (r1 ' + x0).last = r1, r1 = (r ' + x).last, path(r ' + x + H[r1].nxt ', H), path(r ' + x2, H), H[r1].nxt ∈ H, path(r1 ' + x0, H), r1 ∈ H ⊦ H[r1].nxt = null, r1 = r, x2 = [], r1 ∈ x, r1 ∈ x0, dups(x0), r ∈ x2, dups(x2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 75:
((r ' + x).last ' + x0).last = (r ' + x).last, x1 + (r ' + x).last ' = (x + H[(r ' + x).last].nxt ') + r0 ' + y, (r ' + x).last ∈ H, path((r ' + x).last ' + x0, H), H[(r ' + x).last].nxt ∈ H, path(r ' + x1 + (r ' + x).last ', H), path(r ' + x + H[(r ' + x).last].nxt ', H), (r ' + x).last = H[(r ' + x).last].nxt ∨ (r ' + x).last = r0 ∨ (r ' + x).last ∈ y ⊦ (r ' + x).last = r, H[(r ' + x).last].nxt = null, dups(x1), (r ' + x).last ∈ x1, r ∈ x1, dups(x0), (r ' + x).last ∈ x0, (r ' + x).last ∈ x

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ r ∈ r0 ' + x0 ↔ ¬ ¬(r = r0 ∨ r ∈ x0); from specification list[reflist]%in-15: ⊦ x0 + r ' = x1 → r ∈ x1; (forward) from specification list[reflist]%f: ⊦ dups(x0 + r ') ↔ r ∈ x0 ∨ dups(x0); from specification list-dup[reflist]%in-17: ⊦ r ∈ r0 ' ↔ r = r0; from specification list[reflist]%in-append: ⊦ r ∈ x0 + x1 ↔ ¬ ¬(r ∈ x0 ∨ r ∈ x1); from specification list[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%e-02: ⊦ x0 + r ' ≠ []; from specification list[reflist]