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Proof step 13 from proof onlymod-put-01 (previous step)

r -- sk ∈ st0, r ∈ st0 ⊦ sk = _length, sk = _type, (∀ rk. rk ∈ st ↔ rk ∈ st0[r -- sk, val]) ∧ (∀ r0, sk0. r0 ≠ r → st[r0 -- sk0] = st0[r0 -- sk0]) ∧ (r -- _type ∈ st → st[r -- _type] = st0[r -- _type]) ∧ (r -- _length ∈ st → st[r -- _length] = st0[r -- _length]) ↔ (∀ rk. rk ∈ st ↔ rk ∈ st0) ∧ (∀ r0, sk. r0 ≠ r → st[r0 -- sk] = st0[r0 -- sk]) ∧ (r -- _type ∈ st → st[r -- _type] = st0[r -- _type]) ∧ (r -- _length ∈ st → st[r -- _length] = st0[r -- _length])

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 3, ()]

Additional information

[[CONCRETESIMPRULES : ()]]