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Proof step 20 from proof extension (previous step)

0 ≤ i1, 0 < j0, ∀ st0, j0, i0, r0, st, i, r, j1. i→n(j1) < i→n(i1 + 1) → ( true → ( bytevalues(getarray(r, i, j1, st)) = bytevalues(getarray(r0, i0, j0, st0)) ↔ j1 = j0 ∧ (∀ j1. 0 ≤ j1 ∧ j1 < j0 → st[r -- (i + j1) '] .byteval = st0[r0 -- (i0 + j1) '] .byteval) ∧ ¬ j0 < 0 ∨ ¬ 0 < j1 ∧ ¬ 0 < j0)) ⊦ st[r -- i '] .byteval = st0[r0 -- i0 '] .byteval ∧ bytevalues(getarray(r, i + 1, i1, st)) = bytevalues(getarray(r0, i0 + 1, j0 - 1, st0)) ↔ i1 + 1 = j0 ∧ (∀ j. j ≤ i1 ∧ 0 ≤ j → st[r -- (i + j) '] .byteval = st0[r0 -- (i0 + j) '] .byteval)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
j0 = i2 + 1, 0 ≤ i1, 0 < j0, ∀ st0, j0, i0, r0, st, i, r, j1. i→n(j1) < i→n(i1 + 1) → ( true → ( bytevalues(getarray(r, i, j1, st)) = bytevalues(getarray(r0, i0, j0, st0)) ↔ j1 = j0 ∧ (∀ j1. 0 ≤ j1 ∧ j1 < j0 → st[r -- (i + j1) '] .byteval = st0[r0 -- (i0 + j1) '] .byteval) ∧ ¬ j0 < 0 ∨ ¬ 0 < j1 ∧ ¬ 0 < j0)) ⊦ st[r -- i '] .byteval = st0[r0 -- i0 '] .byteval ∧ bytevalues(getarray(r, i + 1, i1, st)) = bytevalues(getarray(r0, i0 + 1, i2, st0)) ↔ i1 + 1 = j0 ∧ (∀ j. j ≤ i1 ∧ 0 ≤ j → st[r -- (i + j) '] .byteval = st0[r0 -- (i0 + j) '] .byteval)

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [j0, 1, i2] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (j0, 1, i2)]]