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Proof step 5 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 6:
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 ⊦ r0 = H[r].nxt, r = null, path(H[r].nxt ' + x, H[r0, ce]) ↔ path(H[r].nxt ' + x, H), 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));%in-rec: ⊦ r ∈ r0 ' + x0 ↔ ¬ ¬(r = r0 ∨ r ∈ x0); from specification list[reflist]%in-06: ⊦ ¬ r ∈ r0 ' + x0 ↔ r ≠ r0 ∧ ¬ r ∈ x0; from specification list[reflist]%At-other: ⊦ r ≠ r0 → H[r0, ce][r] = H[r]; from specification store[linheap]%in-neq: ⊦ r ≠ r0 → (r ∈ H[r0, ce] ↔ r ∈ H); from specification store[linheap]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]