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Proof step 3 from proof disjoint-subset-01 (conclusion)

⊦ ¬ junkref(str, r, st) ∧ ¬ junkref(str, r1, st) ∧ disjoint(str, r ' + r1 ', st) ∧ st ⊆ st1 → disjoint(str, r ' + r1 ', st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
st ⊆ st1, disjoint(str, r ' + r1 ', st) ⊦ junkref(str, r, st), junkref(str, r1, st), disjoint(str, r ' + r1 ', st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]