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Proof step 46 from proof reach-n-path-n (previous step)

# x = 3 + n, path(x, H), ∀ 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) ⊦ ∃ r1. reachable-1(H, hd x, r1) ∧ (∃ r2. reachable-1(H, r1, r2) ∧ reachable-n(H, n, r2, x.last))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 47:
# x = 3 + n, x = r + x1, path(x, H), ∀ 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) ⊦ x = [], ∃ r1. reachable-1(H, r, r1) ∧ (∃ r2. reachable-1(H, r1, r2) ∧ reachable-n(H, n, r2, x.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, r, 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, r, x0