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Proof step 4 from proof reach-n-unique (
previous step
)
reachable-n(H, n0, H[r].nxt, r0), H[r].nxt ∈ H, r ∈ H, reachable-n(H, n0, H[r].nxt, r1), ∀ H, r, r0, r1. reachable-n(H, n0, r, r0) ∧ reachable-n(H, n0, r, r1) → r0 = r1 ⊦ r = null, H[r].nxt = null, r0 = r1
User applied rule all left
And closed the goal!
Rule argument
pos: left 5 terms: H, H[r].nxt, r1, r0 abort: F discard: T computed: T
Additional information
Used terms: H, H[r].nxt, r1, r0