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Proof step 56 from proof reach-n-path-n (previous step)

x2 = [], 1 + 1 + # x2 = 3 + n, ∀ H0, r1, r2. (∃ r3. reachable-1(H0, r1, r3) ∧ reachable-n(H0, n, r3, r2)) ↔ (∃ x. path(x, H0) ∧ hd x = r1 ∧ x.last = r2 ∧ # x = 2 + n), 1 = 1 + # x2 ∧ r ∈ H ∧ H[r].nxt ∈ H ∧ r ≠ null ∧ H[r].nxt ≠ null ∧ H[r].nxt = r4 ∨ 1 ≤ 1 + # x2 ∧ r ∈ H ∧ r ≠ null ∧ H[r].nxt = r4 ∧ path(H[r].nxt ' + x2, H) ⊦ ∃ r0. reachable-1(H, r, r0) ∧ (∃ r1. reachable-1(H, r0, r1) ∧ reachable-n(H, n, r1, (r4 ' + x2).last))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%size-01: ⊦ # [] = 0; from specification list-data[reflist]