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Proof step 6 from proof path-modify (previous step)

x0 = r1 + x, ∀ H, r0, ce. (path(x0, H[r0, ce]) ↔ path(x0, H)) ∨ r0 ∈ x0 ⊦ path(r ' + x0, H[r0, ce]) ↔ path(r ' + x0, H), r0 = r, r0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
r ∈ H, ∀ H0, r2, ce0. (path(H[r].nxt ' + x, H0[r2, ce0]) ↔ path(H[r].nxt ' + x, H0)) ∨ r2 = H[r].nxt ∨ r2 ∈ x ⊦ path(H[r].nxt ' + x, H[r0, ce]) ↔ path(H[r].nxt ' + x, H), r = null, r0 = H[r].nxt, r0 = r, 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);% ⊦ r ∈ r0 ' ↔ r = r0; (s)% ⊦ r ∈ x0 + x1 ↔ r ∈ x0 ∨ r ∈ x1; (s)% ⊦ r ∈ r0 ' + x0 ↔ ¬(r ≠ r0 ∧ ¬ r ∈ x0); (ss)% ⊦ r ≠ r0 → H[r0, ce][r] = H[r]; (s)% ⊦ r ≠ r0 → (r ∈ H[r0, ce] ↔ r ∈ H); (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ r + x0 = r ' + x0; (s)