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Proof step 5 from proof in-1dim-15 (previous step)

i > 0, 0 < i, 0 ≤ j, j < i, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → r -- j ' ∈ addarray(r, val, i0, st)) ⊦ r -- j ' ∈ (addarray(r, val, i - 1, st)[r -- (i - 1) ', val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ j, j < i, 0 < i, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → r -- j ' ∈ addarray(r, val, i0, st)) ⊦ i = j + 1, r -- j ' ∈ addarray(r, val, i - 1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i - j = i0 ↔ i = i0 + j], [CSIMPSEQ : ⊦ i ' = i0 ' ↔ i = i0], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st], [CSIMPSEQ : ⊦ i > j ↔ j < i])]]