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Proof step 21 from proof unchanged-pathex-unchanged (previous step)

path∃(str, r, r0, st), unchanged(str, r, st, st1) ⊦ ∀ r1. (path∃(str, r0, r1, st) ↔ path∃(str, r0, r1, st1)) ∧ (path∃(str, r0, r1, st) → unchangedNorec(r1, st, st1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
unchanged(str, r, st, st1), path∃(str, r, r0, st) ⊦ (path∃(str, r0, r1, st) ↔ path∃(str, r0, r1, st1)) ∧ (path∃(str, r0, r1, st) → unchangedNorec(r1, st, st1))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]