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

∀ 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

User applied rule constructor cut

And obtained the following new goals

Proof step 6:
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

Rule argument

var: x0

Additional information

Used simplifier rules: path-one: ⊦ path(r ', H) ↔ r ∈ H ∧ r ≠ null;%reflist freely generated by [], +;;% ⊦ x0 + [] = x0; (s)%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]% ⊦ r ≠ r0 → (r ∈ H[r0, ce] ↔ r ∈ H); (s)