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Proof step 10 from proof getba-eq (previous step)

0 < i, 0 ≤ i1, 0 ≤ i, 0 < i, ∀ i0. ¬ i0 < i1 ∧ ¬ i1 + i ≤ i0 → st[r -- i0 '] = st[r1 -- i0 '], ∀ r1, r, st, i1, i0. i→n(i0) < i→n(i) → ( 0 ≤ i0 ∧ 0 ≤ i1 ∧ (∀ i2. ¬ i2 < i1 ∧ ¬ i1 + i0 ≤ i2 → st[r -- i2 '] = st[r1 -- i2 ']) → getbytearray(r, i1, i0, st) = getbytearray(r1, i1, i0, st)) ⊦ st[r -- i1 '] .byteval ' + getbytearray(r, i1 + 1, i - 1, st) = st[r1 -- i1 '] .byteval + getbytearray(r1, i1 + 1, i - 1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 ≤ i1, 0 ≤ i, 0 < i, ∀ i0. ¬ i0 < i1 ∧ ¬ i1 + i ≤ i0 → st[r -- i0 '] = st[r1 -- i0 '], ∀ r1, r, st, i1, i0. i→n(i0) < i→n(i) → ( 0 ≤ i0 ∧ 0 ≤ i1 ∧ (∀ i2. ¬ i2 < i1 ∧ ¬ i1 + i0 ≤ i2 → st[r -- i2 '] = st[r1 -- i2 ']) → getbytearray(r, i1, i0, st) = getbytearray(r1, i1, i0, st)) ⊦ st[r -- i1 '] .byteval ' + getbytearray(r, i1 + 1, i - 1, st) = st[r1 -- i1 '] .byteval ' + getbytearray(r1, i1 + 1, i - 1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ by + bytes = by ' + bytes])]]