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Proof step 5 from proof putar-atother (previous step)

i ≠ j ∨ r0 ≠ r, n→i(# vals0) + j < i ∨ i < j, ∀ i0, j0, r1, st0, r2. ¬ j0 ≤ i0 ∨ ¬ i0 < n→i(# vals0) + j0 → (putArray(vals0, r1, j0, st0)[r2 -- i0 ']) = st0[r2 -- i0 '] ⊦ (putArray(vals0, r, j + 1, st)[r0 -- i ']) = st[r0 -- i ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r ≠ r0 → (putArray(vals, r, j, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 4:
∀ i0, j0, r1, st0, r2. ¬ j0 ≤ i0 ∨ ¬ i0 < n→i(# vals0) + j0 → (putArray(vals0, r1, j0, st0)[r2 -- i0 ']) = st0[r2 -- i0 '], n→i(# vals0) + j < i ∨ i < j ⊦ i = j, (putArray(vals0, r, j + 1, st)[r -- i ']) = st[r -- i ']

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]