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Proof step 10 from proof get-alleq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
j ≤ 1, ∀ 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, st) = getarray(r0, i0, j, st0) ↔ (∀ j0. 0 ≤ j0 ∧ j0 < j → st[r -- (i + j0) '] = st0[r0 -- (i0 + j0) '])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i])]]