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Proof step 4 from proof to-plus-from-inf-simp (previous step)

⊦ ⌜λ n0. (n0 < #(nf to n) ⊃ (nf to n)[n0];((nf from n))(n0 - #(nf to n)))⌝ = ⌜nf⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (n0 < n ⊃ nf(n0);nf((n0 - n) + n)) = nf(n0)

Rule argument

Additional information

Used simplifier rules: size-inf-to: ⊦ #(nf to n) = n;%get-infto: m < n ⊦ (nf to n)[m] = nf(m);%get-from-inf: ⊦ ((nf from n))(m) = nf(m + n);%inj-01: ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; from specification stream-basic%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat