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Proof step 16 from proof reach-rec-rev (previous step)

∀ a, b. reach(a, b, n0 + 1) → (∃ c. c ∈ succ(a) ∧ reach(c, b, n0)) ⊦ reach(a, b, n0 +1 + 1) → (∃ c. c ∈ succ(a) ∧ reach(c, b, n0 +1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ reach(a, b, n + 1) ↔ (∃ c. reach(a, c, n) ∧ b ∈ succ(c))
Proof step 15:
reach(a, b, 2 + n0), ∀ a0, b0. (∃ c. reach(a0, c, n0) ∧ b0 ∈ succ(c)) → (∃ c. c ∈ succ(a0) ∧ reach(c, b0, n0)) ⊦ ∃ c. c ∈ succ(a) ∧ (∃ c0. reach(c, c0, n0) ∧ b ∈ succ(c0))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m +1 = m + 1])]]