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Proof step 9 from proof intvalues-at (previous step)

∀ m. m < # vals0 → (intvalues(vals0)[m]) = vals0[m] .intval ⊦ m < # (val + vals0) → (intvalues(val + vals0)[m]) = (val + vals0)[m] .intval

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ intvalues(val ' + vals) = val .intval + intvalues(vals)
Proof step 8:
m ≤ # vals0, ∀ m0. ¬ # vals0 ≤ m0 → (intvalues(vals0)[m0]) = vals0[m0] .intval ⊦ (val .intval ' + intvalues(vals0))[m] = (val ' + vals0)[m] .intval

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ # val ' = 0 +1], [CSIMPSEQ : ⊦ # (vals + vals0) = # vals + # vals0], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ i + ints = i ' + ints], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]