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Proof step 28 from proof frommodified (previous step)

path(ce.nxt ' + x, H[r, ce]), x.last ∈ H, ce.nxt ∈ H ⊦ ce.nxt = null, r = x.last, r = null, x = [], ∃ x0. path(r ' + x0, H[r, ce]) ∧ (r ' + x0).last = x.last

User applied rule exists right

And closed the goal!

Rule argument

pos: right 5 terms: ce.nxt + x abort: F discard: T computed: F

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 + x0 = r ' + x0; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ r ∈ H[r, ce]; (s)% ⊦ H[r, ce][r] = ce; (s)% ⊦ (r ' + r0 ' + x0).last = (r0 ' + x0).last; (s)% ⊦ x0 ≠ [] → (r ' + x0).last = x0.last; (s) Used terms: ce.nxt + x