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

path((r0 ' + x0) + r ' + y, H) ⊦ path(r ' + y, H), path(x0 + r ' + y, H)

User applied rule constructor cut

And closed the goal!

Rule argument

var: x0

Additional information

Used simplifier rules: path-cons: ⊦ path(r ' + r0 ' + x, H) ↔ ¬ ¬(r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(r0 ' + x, H));%path-cons: ⊦ path(r ' + r0 ' + x, H) ↔ ¬ ¬(r ∈ H ∧ r ≠ null ∧ H[r].nxt = r0 ∧ path(r0 ' + x, H));%reflist freely generated by [], +;;%c: ⊦ r + x0 = r ' + x0; from specification list[reflist]%a: ⊦ (x0 + x1) + x2 = x0 + x1 + x2; from specification list[reflist]%Nil: ⊦ [] + x0 = x0; from specification list[reflist]