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Proof step 18 from proof putarray-subset-new-01 (previous step)

⊦ (∀ rk. rk ∈ st → rk ∈ putArray(vals, r, i, st0) ∧ st[rk] = (putArray(vals, r, i, st0)[rk])) ↔ st ⊆ st0, r ∈ st

User applied rule case distinction

And obtained the following new goals

Proof step 11:
∀ rk. rk ∈ st → rk ∈ putArray(vals, r, i, st0) ∧ st[rk] = (putArray(vals, r, i, st0)[rk]) ⊦ st ⊆ st0, r ∈ st
Proof step 17:
st ⊆ st0 ⊦ ∀ rk. rk ∈ st → rk ∈ putArray(vals, r, i, st0) ∧ st[rk] = (putArray(vals, r, i, st0)[rk]), r ∈ st

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]