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Proof step 20 from proof get-alleq (previous step)

getarray(r, i, i1, st) = getarray(r0, i0, i1, st0) ↔ (∀ j. ¬ j < 0 ∧ ¬ i1 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) ']), 0 < i1, ∀ j0, r, i, st, r0, i0, st0. i→n(j0) < i→n(i1 + 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) ']))) ⊦ getarray(r, i, i1, st) = getarray(r0, i0, i1, st0) ∧ st[r -- (i + i1) '] = st0[r0 -- (i0 + i1) '] ↔ (∀ j. j ≤ i1 ∧ 0 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) '])

User applied rule case distinction

And obtained the following new goals

Proof step 13:
getarray(r, i, i1, st) = getarray(r0, i0, i1, st0) ∧ st[r -- (i + i1) '] = st0[r0 -- (i0 + i1) '], getarray(r, i, i1, st) = getarray(r0, i0, i1, st0) ↔ (∀ j. ¬ j < 0 ∧ ¬ i1 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) ']), 0 < i1, ∀ j0, r, i, st, r0, i0, st0. i→n(j0) < i→n(i1 + 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) ']))) ⊦ ∀ j. j ≤ i1 ∧ 0 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) ']
Proof step 19:
∀ j. j ≤ i1 ∧ 0 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) '], getarray(r, i, i1, st) = getarray(r0, i0, i1, st0) ↔ (∀ j. ¬ j < 0 ∧ ¬ i1 ≤ j → st[r -- (i + j) '] = st0[r0 -- (i0 + j) ']), 0 < i1, ∀ j0, r, i, st, r0, i0, st0. i→n(j0) < i→n(i1 + 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) ']))) ⊦ getarray(r, i, i1, st) = getarray(r0, i0, i1, st0) ∧ st[r -- (i + i1) '] = st0[r0 -- (i0 + i1) ']

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]