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Proof step 5 from proof linearity-path (previous step)

y = r1 + y0, path(r ' + y, H), r ∈ H ⊦ [] = y, r = null, ∃ r0, x. y = r0 ' + x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
path(H[r].nxt ' + y0, H), r ∈ H ⊦ r = null, ∃ r0, x. H[r].nxt ' + y0 = r0 ' + x

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]%e-01: ⊦ r ' + x0 ≠ []; from specification list[reflist]%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]