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Proof step 6 from proof path-equiv-reach (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x = 1, path(x) ⊦ x .first = x .last

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]