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Proof step 57 from proof path-equiv-reach (previous step)

∃ x. path(x) ∧ # x = 2 + n0 ∧ x .first = a ∧ x .last = b, ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ ∃ c. reach(a, c, n0) ∧ b ∈ succ(c)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
# x = 2 + n0, path(x), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ ∃ c. reach(x .first, c, n0) ∧ x .last ∈ succ(c)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]