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Proof step 6 from proof putar-atother (previous step)

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 '] ⊦ (((i = j ∧ r0 = r) ⊃ val; putArray(vals0, r, j + 1, st)[r0 -- i '])) = st[r0 -- i ']

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
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 ']

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j < j ↔ i < 0], [CSIMPSEQ : ⊦ ¬ n→i(n) < 0], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ ¬ i < i])]]