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Proof step 15 from proof reach-n-path-n (previous step)

n0 = n +1, n0 ≠ 0 → (∀ H, r, r0. reachable-n(H, n0, r, r0) ↔ (∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = n0 + 1)) ⊦ (∃ r1. reachable-1(H, r, r1) ∧ reachable-n(H, n0, r1, r0)) ↔ (∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = 2 + n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
∀ 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, r, r1) ∧ (∃ r2. reachable-1(H, r1, r2) ∧ reachable-n(H, n, r2, r0))) ↔ (∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = 3 + n)

Rule argument

Additional information

Used simplifier rules: reachable-n-rec: ⊦ reachable-n(H, n + 1, r, r0) ↔ (∃ r1. reachable-1(H, r, r1) ∧ reachable-n(H, n, r1, r0));%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2