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Proof step 6 from proof get-length (previous step)

1 < j, 0 < j, ∀ j0, r, i, st. i→n(j0) < i→n(j) → (0 < j0 → # getarray(r, i, j0, st) = i→n(j0)) ⊦ # (getarray(r, i, j - 1, st) + st[r -- (i + j - 1) ']) = i→n(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 < j, 0 < j, ∀ j0, r, i, st. i→n(j0) < i→n(j) → (0 < j0 → # getarray(r, i, j0, st) = i→n(j0)) ⊦ # getarray(r, i, j - 1, st) + 1 = i→n(j)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ # val ' = 0 +1], [CSIMPSEQ : ⊦ # (vals + vals0) = # vals + # vals0], [CSIMPSEQ : ⊦ vals + val = vals + val '])]]