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Proof step 7 from proof intvalues-at (previous step)

val .intval ' + intvalues(vals0) ≠ [] ∧ m ≠ 0, m ≤ # vals0, ∀ m0. ¬ # vals0 ≤ m0 → (intvalues(vals0)[m0]) = vals0[m0] .intval ⊦ ((val .intval ' + intvalues(vals0)) .rest)[m - 1] = (val ' + vals0)[m] .intval

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m ≤ # vals0, ∀ m0. ¬ # vals0 ≤ m0 → (intvalues(vals0)[m0]) = vals0[m0] .intval ⊦ m = 0, (intvalues(vals0)[m - 1]) = (val ' + vals0)[m] .intval

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ' + ints ≠ []], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ (i ' + ints) .rest = ints])]]