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Proof step 214 from proof pseq-subset (previous step)

st ⊆ st2 ∧ ps=(r, st, r1, st1), ∀ st1, r1, st2, st0, r0, str0. maxpath(str0, r0, st0) < maxpath(str, r, st) → ( st0 ⊆ st2 ∧ ps=(r0, st0, r1, st1) → junkref(str0, r0, st0) ∨ cyclic(str0, r0, st0) ∨ junkref(str0, r1, st1) ∨ cyclic(str0, r1, st1) ∨ ps=(r0, st2, r1, st1)) ⊦ junkref(str, r, st) ∨ cyclic(str, r, st) ∨ junkref(str, r1, st1) ∨ cyclic(str, r1, st1) ∨ ps=(r, st2, r1, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 213:
ps=(r, st, r1, st1), st ⊆ st2, ∀ st1, r1, st2, st0, r0, str0. maxpath(str0, r0, st0) < maxpath(str, r, st) → ( st0 ⊆ st2 ∧ ps=(r0, st0, r1, st1) → junkref(str0, r0, st0) ∨ cyclic(str0, r0, st0) ∨ junkref(str0, r1, st1) ∨ cyclic(str0, r1, st1) ∨ ps=(r0, st2, r1, st1)) ⊦ ps=(r, st2, r1, st1), cyclic(str, r1, st1), junkref(str, r1, st1), cyclic(str, r, st), junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]