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Proof step 58 from proof path-equiv-reach (previous step)

∀ 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)) ↔ (∃ x. path(x) ∧ # x = 2 + n0 ∧ x .first = a ∧ x .last = b)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 25:
∃ c. reach(a, c, n0) ∧ b ∈ succ(c), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ ∃ x. path(x) ∧ # x = 2 + n0 ∧ x .first = a ∧ x .last = b
Proof step 57:
∃ 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)

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]