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Proof step 7 from proof boolvals-get (previous step)

n ≠ 0, n ≤ # vals0, ∀ n. n < # vals0 → (booleanvalues(vals0)[n] ↔ vals0[n] .boolval) ⊦ (val .boolval ' + booleanvalues(vals0))[n] ↔ vals0[n -1] .boolval

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ≤ # vals0, ∀ n. n < # vals0 → (booleanvalues(vals0)[n] ↔ vals0[n] .boolval) ⊦ n = 0, (val .boolval ' + booleanvalues(vals0))[n] ↔ vals0[n - 1] .boolval

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : nat, , pred-03, ⊦ n -1 = n - 1])]]