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Proof step 10 from proof linearity-path (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
path(H[r].nxt ' + x0, H), r ∈ H, ∀ r, H, y. path(r ' + x0, H) ∧ path(r ' + y, H) → x0 = y ∨ (∃ r0, x1. y = x0 + r0 ' + x1) ∨ (∃ r0, y0. x0 = y + r0 ' + y0) ⊦ r = null, ∃ r0, y. H[r].nxt ' + x0 = r0 ' + y

Rule argument

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ r ∈ H ∧ r ≠ null;% ⊦ [] + x0 = x0; (s)% ⊦ x0 + [] = x0; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ r ' + x0 ≠ []; (s)