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Proof step 6 from proof getarray-union (previous step)

0 < i0 + 1, 0 ≤ i0, ∀ st0, st, i, r, j0. i→n(j0) < i→n(i0 + 1) → (true → getarray(r, i, j0, st ∪ st0) = getarray(r, i, j0, st0) ∨ r ∈ st) ⊦ st0[r -- i '] ' + getarray(r, i + 1, i0, st ∪ st0) = st0[r -- i '] ' + getarray(r, i + 1, i0 + 1 - 1, st0), r -- i ' ∈ st, r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i0, ∀ st0, st, i, r, j0. i→n(j0) < i→n(i0 + 1) → (true → getarray(r, i, j0, st ∪ st0) = getarray(r, i, j0, st0) ∨ r ∈ st) ⊦ getarray(r, i + 1, i0, st ∪ st0) = getarray(r, i + 1, i0, st0), r -- i ' ∈ st, r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j], [CSIMPSEQ : ⊦ (~1) < i ↔ 0 ≤ i], [CSIMPSEQ : ⊦ ((* j)) < i0 + ((* j0)) ↔ j - j0 < i0], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ vals + vals0 = vals + vals1 ↔ vals0 = vals1], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + 0 = i], [CSIMPSEQ : ⊦ ((* i)) + i0 - ((* j)) = (i - j) + i0])]]