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Proof step 72 from proof disjoint-nojunk-newref-subset (conclusion)

⊦ st ⊆ st1 ∧ ¬ junkref(str, r1, st) ∧ newref(r, st) ∧ (∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref) → disjoint(str, r ' + r1 ', st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 71:
st ⊆ st1, newref(r, st), ∀ r0. path∃(str, r, r0, st1) → newref(r0, st) ∨ r0 = jvmref ⊦ r = jvmref, disjoint(str, r ' + r1 ', st1), junkref(str, r1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ newref(r, st) → r ≠ jvmref])]]