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Proof step 5 from proof fits-subset-01 (previous step)

st ⊆ st0 ⊦ fits(r, sks, st0) ↔ fits(r, sks, st), junkref(str, r, st)

User applied rule case distinction

And obtained the following new goals

Proof step 2:
fits(r, sks, st0), st ⊆ st0 ⊦ fits(r, sks, st), junkref(str, r, st)
Proof step 4:
fits(r, sks, st), st ⊆ st0 ⊦ fits(r, sks, st0), junkref(str, r, st)

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]