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Proof step 4 from proof path-split-last (previous step)

r ∈ H, ∀ r, r0, H. path(r ' + x0 + r0 ', H) ↔ path(r ' + x0, H) ∧ H[(r ' + x0).last].nxt = r0 ∧ H[(r ' + x0).last].nxt ∈ H ∧ H[(r ' + x0).last].nxt ≠ null ⊦ r = null, path(H[r].nxt ' + x0 + r0 ', H) ↔ path(H[r].nxt ' + x0, H) ∧ H[(H[r].nxt ' + x0).last].nxt = r0 ∧ H[(H[r].nxt ' + x0).last].nxt ∈ H ∧ H[(H[r].nxt ' + x0).last].nxt ≠ null

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: H[r].nxt, r0, H abort: F discard: T computed: T

Additional information

Used simplifier rules: a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist] Used terms: H[r].nxt, r0, H