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Proof step 18 from proof unchanged-levelsubset-equiv (previous step)

¬ ¬ (∀ r0. (path∃(str0, r, r0, st) ↔ path∃(str0, r, r0, st0)) ∧ (path∃(str0, r, r0, st) → unchangedNorec(r0, st, st0))), level⊆(str, str0), unchanged(str0, r, st, st0) ⊦ junkref(str, r, st), (path∃(str, r, r0, st) ↔ path∃(str, r, r0, st0)) ∧ (path∃(str, r, r0, st) → unchangedNorec(r0, st, st0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
unchanged(str0, r, st, st0), level⊆(str, str0), ∀ r0. (path∃(str0, r, r0, st) ↔ path∃(str0, r, r0, st0)) ∧ (path∃(str0, r, r0, st) → unchangedNorec(r0, st, st0)) ⊦ junkref(str, r, st), (path∃(str, r, r0, st) ↔ path∃(str, r, r0, st0)) ∧ (path∃(str, r, r0, st) → unchangedNorec(r0, st, st0))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]