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Proof step 84 from proof rem-stays-reachable-aux2 (previous step)

¬ x3 ≠ [], (r ' + x1).last = (r ' + x).last, x1 + r2 ' + x3 = x + H[(r2 ' + x3).last].nxt ' + r0 ' + y, ((r2 ' + x3).last ' + x0).last = (r ' + x).last, (r ' + x).last ∈ H, path(r ' + x1, H), path(r ' + x + H[(r2 ' + x3).last].nxt ', H), path(r ' + x1 + r2 ' + x3, H), H[(r2 ' + x3).last].nxt ∈ H, (r2 ' + x3).last ∈ H, path((r2 ' + x3).last ' + x0, H) ⊦ H[(r2 ' + x3).last].nxt = null, (r2 ' + x3).last = r, r = r2, (r2 ' + x3).last ∈ x, (r2 ' + x3).last ∈ x0, dups(x0), r ∈ x1, r ∈ x3, r2 ∈ x1, r2 ∈ x3, dups(x1 + x3), dups(x1), dups(x3)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 85:
(r2 ' + x0).last = (r ' + x).last, x1 + r2 ' = x + H[r2].nxt ' + r0 ' + y, (r ' + x1).last = (r ' + x).last, path(r2 ' + x0, H), r2 ∈ H, H[r2].nxt ∈ H, path(r ' + x1 + r2 ', H), path(r ' + x + H[r2].nxt ', H), path(r ' + x1, H), (r ' + x).last ∈ H, r2 = H[r2].nxt ∨ r2 = r0 ∨ r2 ∈ y ⊦ H[r2].nxt = null, r = r2, r2 ∈ x1, r ∈ x1, dups(x0), r2 ∈ x0, r2 ∈ x, dups(x1)

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ r ∈ r0 ' + x0 ↔ ¬ ¬(r = r0 ∨ r ∈ x0); from specification list[reflist]%in-append: ⊦ r ∈ x0 + x1 ↔ ¬ ¬(r ∈ x0 ∨ r ∈ x1); from specification list[reflist]%in-15: ⊦ x0 + r ' = x1 → r ∈ x1; (forward) from specification list[reflist]%e-01: ⊦ ¬ dups([]); from specification list-dup[reflist]%in-empty: ⊦ ¬ r ∈ []; from specification list[reflist]%l: ⊦ r '.last = r; from specification list-last[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%app: ⊦ x0 + [] = x0; from specification list[reflist]