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Proof step 19 from proof reach-n-path-n (previous step)

n = 0, reachable-1(H, r, r1), reachable-1(H, r1, r2), reachable-n(H, n, r2, r0), ∀ 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. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = 3 + n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
reachable-1(H, r1, r0), reachable-1(H, r, r1), ∀ H0, r4, r5. reachable-1(H0, r4, r5) ↔ (∃ x. path(x, H0) ∧ hd x = r4 ∧ x.last = r5 ∧ # x = 2) ⊦ ∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = 3

Rule argument

Additional information

Used simplifier rules: reachable-n-base: ⊦ reachable-n(H, 0, r, r0) ↔ r = r0;