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Proof step 6 from proof contractinf-from1 (previous step)

¬ ⌜n⌝ < # ⌜nf(0) '⌝ ⊦ (⌜nf(0) '⌝ + ⌜nf from 1⌝ )[n] = nf(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ nf((n - 1) + 1) = nf(n), n = 0

Rule argument

Additional information

Used simplifier rules: get-append-notls: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n];%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%sel-inf: ⊦ ⌜nf⌝ [n] = nf(n);%get-from-inf: ⊦ ((nf from n))(m) = nf(m + n);%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-basic%len-02: ⊦ # a ' = 0 +1; from specification list