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Proof step 5 from proof getarray-superset (conclusion)

⊦ st ⊆ st0 ∧ is_arrayref(r, i + j, st) ∧ 0 ≤ i → getarray(r, i, j, st0) = getarray(r, i, j, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st ⊆ st0 ∧ is_arrayref(r, i + j, st) ∧ 0 ≤ i → getarray(r, i, j, st) = getarray(r, i, j, st0)
Proof step 4:
¬ (st ⊆ st0 ∧ is_arrayref(r, i + j, st) ∧ 0 ≤ i) ⊦ st ⊆ st0 ∧ is_arrayref(r, i + j, st) ∧ 0 ≤ i → getarray(r, i, j, st0) = getarray(r, i, j, st)

Rule argument

[crewrite-arg: getarray-subset , F , [st, st0, r, i, j] → [st, st0, r, i, j] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, st0, r, i, j)]]