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Proof step 6 from proof putarray-subset-new-01 (previous step)

st[r0 -- sk] = (putArray(vals, r, i, st0)[r0 -- sk]), r0 -- sk ∈ st, r0 -- sk ∈ putArray(vals, r, i, st0) ⊦ r ∈ st, r0 -- sk ∈ st0 ∧ st[r0 -- sk] = st0[r0 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
st[r0 -- sk] = (putArray(vals, r, i, st0)[r0 -- sk]), r0 -- sk ∈ st, r0 -- sk ∈ putArray(vals, r, i, st0), r0 ∈ st, r0 ∈ putArray(vals, r, i, st0) ⊦ r ∈ st, r0 -- sk ∈ st0 ∧ (putArray(vals, r, i, st0)[r0 -- sk]) = st0[r0 -- sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]