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Proof step 4 from proof reachable-n-trans (previous step)

H[r].nxt ∈ H, r ∈ H, reachable-n(H, n0, r0, r1), reachable-n(H, n1, H[r].nxt, r0), ∀ H, r, r0, n0, r1. reachable-n(H, n1, r, r0) ∧ reachable-n(H, n0, r0, r1) → reachable-n(H, n1 + n0, r, r1) ⊦ r = null, H[r].nxt = 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 5:
H[r].nxt ∈ H, r ∈ H, reachable-n(H, n0, r0, r1), reachable-n(H, n1, H[r].nxt, r0), reachable-n(H, n1, H[r].nxt, r0) ∧ reachable-n(H, n0, r0, r1) → reachable-n(H, n1 + n0, H[r].nxt, r1) ⊦ r = null, H[r].nxt = null, ∃ r2. reachable-1(H, r, r2) ∧ reachable-n(H, n1 + n0, r2, r1)

Rule argument

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

Additional information

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