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Proof step 8 from proof path-equiv-reach (previous step)

⊦ reach(a, b, 0) ↔ (∃ x. path(x) ∧ # x = 0 + 1 ∧ x .first = a ∧ x .last = b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ reach(a, b, 0) ↔ a = b
Proof step 7:
⊦ a = b ↔ (∃ x. path(x) ∧ # x = 1 ∧ x .first = a ∧ x .last = b)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]