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Proof step 49 from proof reach-n-path-n (previous step)

1 + # x1 = 3 + n, x1 = [], path(r ' + x1, 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) ⊦ ∃ r0. reachable-1(H, r, r0) ∧ (∃ r1. reachable-1(H, r0, r1) ∧ reachable-n(H, n, r1, (r ' + x1).last))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # [] = 0; (s)