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Proof step 5 from proof reachable-n-trans (previous step)

reachable-n(H, n0, r0, r1), reachable-n(H, n1, H[r].nxt, r0), H[r].nxt ∈ H, r ∈ H, ∀ H, n0, r0, r1, r. reachable-n(H, n0, r0, r1) ∧ reachable-n(H, n1, r, r0) → reachable-n(H, n1 + n0, r, r1) ⊦ H[r].nxt = null, r = null, ∃ r2. reachable-1(H, r, r2) ∧ reachable-n(H, n1 + n0, r2, r1)

User applied rule all left

And obtained the following new goals

Proof step 6:
reachable-n(H, n1, H[r].nxt, r0), reachable-n(H, n0, r0, r1), r ∈ H, H[r].nxt ∈ H, reachable-n(H, n0, r0, r1) ∧ reachable-n(H, n1, H[r].nxt, r0) → reachable-n(H, n1 + n0, H[r].nxt, r1) ⊦ H[r].nxt = null, r = null, ∃ r2. reachable-1(H, r, r2) ∧ reachable-n(H, n1 + n0, r2, r1)

Rule argument

pos: left 5 terms: H, n0, r0, r1, H[r].nxt abort: F discard: T computed: T

Additional information

Used terms: H, n0, r0, r1, H[r].nxt