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Proof step 7 from proof path-equiv-reach (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a = b ⊦ a = b ↔ (∃ x. path(x) ∧ # x = 1 ∧ x .first = a ∧ x .last = b)
Proof step 6:
⊦ a = b, a = b ↔ (∃ x. path(x) ∧ # x = 1 ∧ x .first = a ∧ x .last = b)

Rule argument

[fma-arg: a = b]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (a = b)]]