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Proof step 29 from proof getarray-put=set (previous step)

0 ≤ i1, i1 ≤ i0, 0 ≤ i0, 0 < i0, ∀ val, st, i, r, i1, j0. i→n(j0) < i→n(i0 + 1) → (0 < j0 ∧ i1 < j0 ∧ 0 ≤ i1 → getarray(r, i, j0, st[r -- (i1 + i) ', val]) = (getarray(r, i, j0, st)[i→n(i1), val])) ⊦ getarray(r, i, i0, st[r -- (i1 + i) ', val]) + ((i0 = i1 ⊃ val; st[r -- (i + i0) '])) ' = (getarray(r, i, i0 + 1, st)[i→n(i1), val])

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 15:
i0 = i1, 0 ≤ i1, i1 ≤ i0, 0 ≤ i0, 0 < i0, ∀ val, st, i, r, i1, j0. i→n(j0) < i→n(i0 + 1) → (0 < j0 ∧ i1 < j0 ∧ 0 ≤ i1 → getarray(r, i, j0, st[r -- (i1 + i) ', val]) = (getarray(r, i, j0, st)[i→n(i1), val])) ⊦ getarray(r, i, i0, st[r -- (i1 + i) ', val]) + val ' = (getarray(r, i, i0 + 1, st)[i→n(i1), val])
Proof step 28:
i0 ≠ i1, 0 ≤ i1, i1 ≤ i0, 0 ≤ i0, 0 < i0, ∀ val, st, i, r, i1, j0. i→n(j0) < i→n(i0 + 1) → (0 < j0 ∧ i1 < j0 ∧ 0 ≤ i1 → getarray(r, i, j0, st[r -- (i1 + i) ', val]) = (getarray(r, i, j0, st)[i→n(i1), val])) ⊦ getarray(r, i, i0, st[r -- (i1 + i) ', val]) + st[r -- (i + i0) '] ' = (getarray(r, i, i0 + 1, st)[i→n(i1), val])

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]