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Proof step 23 from proof frommodified (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: ce.nxt ' abort: F discard: T computed: F

Additional information

Used simplifier rules: path-two: ⊦ path(r ' + r0 ', H) ↔ ¬ ¬(r ∈ H ∧ r0 ∈ H ∧ r ≠ null ∧ r0 ≠ null ∧ H[r].nxt = r0);%insin: ⊦ r ∈ H[r, ce]; from specification store[linheap]%in-neq: ⊦ r ≠ r0 → (r ∈ H[r0, ce] ↔ r ∈ H); from specification store[linheap]%At-same: ⊦ H[r, ce][r] = ce; from specification store[linheap]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%last-def: ⊦ (x0 + r ').last = r; from specification list-last[reflist] Used terms: ce.nxt '