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Proof step 105 from proof rem-stays-reachable-aux2 (previous step)

x1 + r2 ' + x3 = x + H[r1].nxt ' + r0 ' + y, (r1 ' + x0).last = (r ' + x).last, (r ' + x1).last = (r ' + x).last, x3 = x2 + r1 ', path(r1 ' + x0, H), path(r ' + x + H[r1].nxt ', H), path(r ' + x1 + r2 ' + x3, H), path(r ' + x1, H), H[r1].nxt ∈ H, r1 ∈ H, (r ' + x).last ∈ H ⊦ H[r1].nxt = null, r = r2, r1 = r, x3 = [], dups(x3), dups(x1), dups(x1 + x3), dups(x0), r2 ∈ x3, r2 ∈ x1, r1 ∈ x, r1 ∈ x0, r ∈ x1, r ∈ x3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 106:
x1 + r2 ' + x2 + r1 ' = x + H[r1].nxt ' + r0 ' + y, (r1 ' + x0).last = (r ' + x).last, (r ' + x1).last = (r ' + x).last, path(r1 ' + x0, H), path(r ' + x + H[r1].nxt ', H), path(r ' + x1 + r2 ' + x2 + r1 ', H), path(r ' + x1, H), H[r1].nxt ∈ H, r1 ∈ H, (r ' + x).last ∈ H, r1 = H[r1].nxt ∨ r1 = r0 ∨ r1 ∈ y ⊦ H[r1].nxt = null, r = r2, r1 = r, r2 = r1, dups(x2), dups(x1), dups(x1 + x2), dups(x0), r1 ∈ x2, r1 ∈ x1, r2 ∈ x2, r2 ∈ x1, r1 ∈ x, r1 ∈ x0, r ∈ x1, r ∈ x2

Rule argument

Additional information

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