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Proof step 6 from proof get-alleq (previous step)

∀ j0, r, i, st, r0, i0, st0. i→n(j0) < i→n(1) → (true → (getarray(r, i, j0, st) = getarray(r0, i0, j0, st0) ↔ (∀ j1. 0 ≤ j1 ∧ j1 < j0 → st[r -- (i + j1) '] = st0[r0 -- (i0 + j1) ']))) ⊦ st[r -- i '] = st0[r0 -- i0 '] ↔ (∀ j. j ≤ 0 ∧ 0 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) '])

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∀ j. j ≤ 0 ∧ 0 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) '], ∀ j0, r, i, st, r0, i0, st0. i→n(j0) < i→n(1) → (true → (getarray(r, i, j0, st) = getarray(r0, i0, j0, st0) ↔ (∀ j1. 0 ≤ j1 ∧ j1 < j0 → st[r -- (i + j1) '] = st0[r0 -- (i0 + j1) ']))) ⊦ st[r -- i '] = st0[r0 -- i0 ']

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ≤ j → (j ≤ i ↔ i = j)], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ i + 0 = i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i])]]