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Proof step 7 from proof linearity-path (previous step)

∀ r, H, y. path(r ' + x0, H) ∧ path(r ' + y, H) → x0 = y ∨ (∃ r0, y0. x0 = y + r0 ' + y0) ∨ (∃ r0, x1. y = x0 + r0 ' + x1), path(r ' + (r0 + x0), H), path(r ' + y, H) ⊦ r0 + x0 = y, ∃ r, y0. r0 + x0 = y + r ' + y0, ∃ r, x1. y = (r0 + x0) + r ' + x1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
path(H[r].nxt ' + x0, H), r ∈ H, path(r ' + y, H), ∀ r, H, y. path(r ' + x0, H) ∧ path(r ' + y, H) → x0 = y ∨ (∃ r0, y0. x0 = y + r0 ' + y0) ∨ (∃ r0, x1. y = x0 + r0 ' + x1) ⊦ r = null, H[r].nxt ' + x0 = y, ∃ r1, x. y = (H[r].nxt ' + x0) + r1 ' + x, ∃ r1, y0. H[r].nxt ' + x0 = y + r1 ' + y0

Rule argument

Additional information

Used simplifier rules: path-cons: ⊦ path(r ' + r0 ' + x, H) ↔ ¬ ¬(r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(r0 ' + x, H));%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]