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Proof step 10 from proof append-cancel-l (previous step)

⌜λ n. (n < # x ⊃ x[n];nf(n - # x))⌝ = ⌜λ n. (n < # x ⊃ x[n];nf0(n - # x))⌝ ⊦ nf = nf0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(λ n. (n < # x ⊃ x[n];nf(n - # x))) = (λ n. (n < # x ⊃ x[n];nf0(n - # x))) ⊦ nf = nf0

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)