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Proof step 3 from proof reach-n-path-n (previous step)

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)

User applied rule constructor cut

And obtained the following new goals

Proof step 4:
n0 = 0, 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)
Proof step 15:
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)

Rule argument

var: n0

Additional information

Used simplifier rules: nat freely generated by 0, +1;;