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Proof step 10 from proof reach-n-path-n (previous step)

r ∈ H, H[r].nxt ∈ H ⊦ H[r].nxt = null, r = null, ∃ x. path(x, H) ∧ hd x = r ∧ x.last = H[hd x].nxt ∧ # x = 2

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: r ' + H[r].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);%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%c-08: ⊦ hd(r ' + x0) = r; from specification list[reflist]%last-def: ⊦ (x0 + r ').last = r; from specification list-last[reflist]%len: ⊦ #(x0 + x1) = # x0 + # x1; from specification list[reflist]%len-02: ⊦ # r ' = 0 +1; from specification list[reflist] Used terms: r ' + H[r].nxt '