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Proof step 113 from proof rem-stays-reachable-aux2 (previous step)

x1 + r2 ' + x2 ≠ [], (r0 ' + x0).last = (r ' + x).last, x = (x1 + r2 ' + x2).butlast ∧ H[r0].nxt = (x1 + r2 ' + x2).last, (r ' + x1).last = (r ' + x).last, path(r0 ' + x0, H), r0 ∈ H, H[r0].nxt ∈ H, path(r ' + x1 + r2 ' + x2 + r0 ', H), path(r ' + x1 + r2 ' + x2, H), path(r ' + x1, H), (r ' + x).last ∈ H, H[r0].nxt ∈ x1 ∨ H[r0].nxt = r2 ∨ H[r0].nxt ∈ x2 ⊦ r2 = r0, r = r2, r0 = r, H[r0].nxt = null, r0 ∈ x2, r0 ∈ x1, r2 ∈ x2, r2 ∈ x1, r ∈ x2, r ∈ x1, dups(x0), r0 ∈ x0, r0 ∈ x, dups(x2), dups(x1), dups(x1 + x2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 114:
(r0 ' + x0).last = (r ' + x1).last, H[r0].nxt = (r2 ' + x2).last, (r ' + x1).last = (r ' + x1 + (r2 ' + x2).butlast).last, path(r0 ' + x0, H), r0 ∈ H, H[r0].nxt ∈ H, path(r ' + x1 + r2 ' + x2 + r0 ', H), path(r ' + x1 + r2 ' + x2, H), path(r ' + x1, H), (r ' + x1).last ∈ H, H[r0].nxt ∈ x1 ∨ H[r0].nxt = r2 ∨ H[r0].nxt ∈ x2 ⊦ r2 = r0, r = r2, r0 = r, H[r0].nxt = null, dups(x2), dups(x1), dups(x1 + x2), r0 ∈ x2, r0 ∈ x1, r2 ∈ x2, r2 ∈ x1, r ∈ x2, r ∈ x1, dups(x0), r0 ∈ x0, r0 ∈ (r2 ' + x2).butlast

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ r ∈ x0 + x1 ↔ ¬ ¬(r ∈ x0 ∨ r ∈ x1); from specification list[reflist]%e-03: ⊦ x0 + r ' + x1 ≠ []; from specification list[reflist]%lapp-01: ⊦ (x0 + r ' + x1).last = (r ' + x1).last; from specification list-last[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%bin: ⊦ (x0 + r ' + x1).butlast = x0 + (r ' + x1).butlast; from specification list-last[reflist]