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Proof step 12 from proof comp-same (previous step)

st[r -- (i1 + i + (j0 - 1) * (i1 - i2)) '] = st[r -- (i1 + i) '], i2 < i1, i + i4 * (i1 - i2) < i3, j0 ≤ i4, 0 ≤ i2, 0 < i1, 0 ≤ j0, 0 ≤ i4, 0 ≤ i, ∀ j. ¬ j < 0 ∧ ¬ i3 ≤ j → st[r -- (i1 + j) '] = st[r -- (i2 + j) '], ∀ j, i2, i1, i, i4, i3, st, r. i→n(j) < i→n(j0) → ( i2 < i1 ∧ i + i4 * (i1 - i2) < i3 ∧ 0 ≤ i2 ∧ 0 ≤ i4 ∧ 0 ≤ j ∧ j ≤ i4 ∧ 0 ≤ i ∧ (∀ j. 0 ≤ j ∧ j < i3 → st[r -- (i1 + j) '] = st[r -- (i2 + j) ']) → st[r -- (i1 + i + j * (i1 - i2)) '] = st[r -- (i1 + i) ']) ⊦ st[r -- (i1 + i + j0 * (i1 - i2)) '] = st[r -- (i1 + i) ']

User applied rule all left

And obtained the following new goals

Proof step 11:
st[r -- (i1 + i + (j0 - 1) * (i1 - i2)) '] = st[r -- (i1 + i) '], i2 < i1, i + i4 * (i1 - i2) < i3, j0 ≤ i4, 0 ≤ i2, 0 < i1, 0 ≤ j0, 0 ≤ i4, 0 ≤ i, ¬ i + j0 * (i1 - i2) < 0 ∧ ¬ i3 ≤ i + j0 * (i1 - i2) → st[r -- (i1 + i + j0 * (i1 - i2)) '] = st[r -- (i2 + i + j0 * (i1 - i2)) '], ∀ j. ¬ j < 0 ∧ ¬ i3 ≤ j → st[r -- (i1 + j) '] = st[r -- (i2 + j) '], ∀ j, i2, i1, i, i4, i3, st, r. i→n(j) < i→n(j0) → ( i2 < i1 ∧ i + i4 * (i1 - i2) < i3 ∧ 0 ≤ i2 ∧ 0 ≤ i4 ∧ 0 ≤ j ∧ j ≤ i4 ∧ 0 ≤ i ∧ (∀ j. 0 ≤ j ∧ j < i3 → st[r -- (i1 + j) '] = st[r -- (i2 + j) ']) → st[r -- (i1 + i + j * (i1 - i2)) '] = st[r -- (i1 + i) ']) ⊦ st[r -- (i1 + i + j0 * (i1 - i2)) '] = st[r -- (i1 + i) ']

Rule argument

[EXRARG : pos: left 10, [extquanttermlist: (i + j0 * (i1 - i2)) abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (i + j0 * (i1 - i2))]]