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Proof step 26 from proof get-alleq (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
j = i1 + 1, 1 < j, 0 < j, ∀ j0, r, i, st, r0, i0, st0. i→n(j0) < i→n(j) → (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) + st[r -- (i + j - 1) '] ' = getarray(r0, i0, j, st0) ↔ (∀ j0. 0 ≤ j0 ∧ j0 < j → st[r -- (i + j0) '] = st0[r0 -- (i0 + j0) '])

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [j, 1, i1] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (j, 1, i1)]]