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Proof step 257 from proof cyclic-pseq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 256:
cyclic(str, r, st), ps=(r, st, r1, st1), ∀ st, r, st0, r0, str0. maxpath(str0, r0, st0) < maxpath(str, r1, st1) → (ps=(r, st, r0, st0) ∧ cyclic(str0, r, st) → junkref(str0, r, st) ∨ junkref(str0, r0, st0) ∨ cyclic(str0, r0, st0)) ⊦ cyclic(str, r1, st1), junkref(str, r1, st1), junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]