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Proof step 7 from proof unchanged-symmetric (previous step)

∀ r0. (path∃(str, r, r0, st) ↔ path∃(str, r, r0, st1)) ∧ (path∃(str, r, r0, st) → unchangedNorec(r0, st, st1)) ⊦ ∀ r0. (path∃(str, r, r0, st1) ↔ path∃(str, r, r0, st)) ∧ (path∃(str, r, r0, st1) → unchangedNorec(r0, st1, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ r0. (path∃(str, r, r0, st) ↔ path∃(str, r, r0, st1)) ∧ (path∃(str, r, r0, st) → unchangedNorec(r0, st, st1)) ⊦ (path∃(str, r, r0, st1) ↔ path∃(str, r, r0, st)) ∧ (path∃(str, r, r0, st1) → unchangedNorec(r0, st1, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]