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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
n = m + 1, n ≤ # vals0, ∀ n. n < # vals0 → (booleanvalues(vals0)[n] ↔ vals0[n] .boolval) ⊦ n = 0, (val .boolval ' + booleanvalues(vals0))[n] ↔ vals0[m] .boolval

Rule argument

[x0lemmaarg: spec-name: nat instance-name: lemma-name: elim-pred-c antecedent: n ≠ 0 succedent: m = n - 1 ↔ n = m + 1 substlist: [n, m] → [n, m] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (n, m)]]