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Proof step 63 from proof reach-n-path-n (previous step)

# x = 2 + n, # x2 = n + 1, x.last = x2.last, hd x = H[r].nxt, path(x, H), path((hd x)' + x2, H), reachable-1(H, hd x, r0), reachable-n(H, n, r0, x2.last), r ∈ H ⊦ r = null, x2 = [], ∃ r0. reachable-1(H, r, r0) ∧ (∃ r1. reachable-1(H, r0, r1) ∧ reachable-n(H, n, r1, x2.last))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 64:
# x2 = n + 1, # x = 2 + n, x.last = x2.last, r2 = H[r].nxt, x = r2 + x1, path(r2 ' + x2, H), path(x, H), reachable-1(H, r2, r0), reachable-n(H, n, r0, x2.last), r ∈ H ⊦ r = null, x = [], x2 = [], ∃ r0. reachable-1(H, r, r0) ∧ (∃ r1. reachable-1(H, r0, r1) ∧ reachable-n(H, n, r1, x2.last))

Rule argument

spec: list-data, inst: reflist name: elim-list-c seq: x0 ≠ [] ⊦ r = hd x0 ∧ x1 = tl x0 ↔ x0 = r + x1 subst: [x0, r, x1] → [x, r2, x0]

Additional information

Used simplifier rules: ⊦ # [] = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%x0 ≠ [] ⊦ r = hd x0 ∧ x1 = tl x0 ↔ x0 = r + x1; (elim) Used terms: x, r2, x0