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Proof step 4 from proof reachless-succ-edge (previous step)

reach(source, a, n0), b ∈ succ(a) ⊦ succ(a) = ∅, # succ(a) = 0, n0 + 1 < n0 + 2 ∧ reach(source, b, n0 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ reach(a, b, n + 1) ↔ (∃ c. reach(a, c, n) ∧ b ∈ succ(c))
Proof step 3:
reach(source, a, n0), b ∈ succ(a) ⊦ succ(a) = ∅, # succ(a) = 0, ∃ c. reach(source, c, n0) ∧ b ∈ succ(c)

Rule argument

[emptyarg]

Additional information

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