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Proof step 5 from proof reach-n-split (previous step)

H[r].nxt ∈ H, r ∈ H, reachable-n(H, n0 + m0, H[r].nxt, r0), ∀ H, n0, r, r0. reachable-n(H, n0 + m0, r, r0) → n0 = 0 ∨ (∃ r1. reachable-n(H, m0, r, r1) ∧ reachable-n(H, n0, r1, r0)) ⊦ r = null, H[r].nxt = null, n0 = 0, ∃ r2. (∃ r1. reachable-1(H, r, r1) ∧ reachable-n(H, m0, r1, r2)) ∧ reachable-n(H, n0, r2, r0)

User applied rule all left

And obtained the following new goals

Proof step 6:
H[r].nxt ∈ H, r ∈ H, reachable-n(H, n0 + m0, H[r].nxt, r0), reachable-n(H, n0 + m0, H[r].nxt, r0) → n0 = 0 ∨ (∃ r1. reachable-n(H, m0, H[r].nxt, r1) ∧ reachable-n(H, n0, r1, r0)) ⊦ r = null, H[r].nxt = null, n0 = 0, ∃ r2. (∃ r1. reachable-1(H, r, r1) ∧ reachable-n(H, m0, r1, r2)) ∧ reachable-n(H, n0, r2, r0)

Rule argument

pos: left 4 terms: H, n0, H[r].nxt, r0 abort: F discard: T computed: F

Additional information

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