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Proof step 11 from proof cons-eq (previous step)

(λ n. (n < # a ' ⊃ a '[n];nf(n - # a '))) = (λ n. (n < # a0 ' ⊃ a0 '[n];nf0(n - # a0 '))) ⊦ a = a0 ∧ ⌜nf⌝ = ⌜nf0⌝

User applied rule expand left

And obtained the following new goals

Proof step 12:
(0 < # a ' ⊃ a '[0];nf(0 - # a ')) = (0 < # a0 ' ⊃ a0 '[0];nf0(0 - # a0 ')), (λ n. (n < # a ' ⊃ a '[n];nf(n - # a '))) = (λ n. (n < # a0 ' ⊃ a0 '[n];nf0(n - # a0 '))) ⊦ a = a0 ∧ ⌜nf⌝ = ⌜nf0⌝

Rule argument

pos: left 1 terms: 0 abort: F discard: F computed: F

Additional information

Used terms: 0