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Proof step 6 from proof cons-eq (previous step)

⌜λ n. (n < # a ' ⊃ a '[n];nf(n - # a '))⌝ = ⌜a0 '⌝ + σ0 ⊦ a = a0 ∧ ⌜nf⌝ = σ0

User applied rule structural induction

And obtained the following new goals

Proof step 7:
⌜λ n. (n < # a ' ⊃ a '[n];nf(n - # a '))⌝ = ⌜a0 '⌝ + ⌜nf0⌝ ⊦ a = a0 ∧ ⌜nf⌝ = ⌜nf0⌝

Rule argument

indvar: σ0

Additional information

Used simplifier rules: neq-fin-inf-03: ⊦ σ0 + ⌜x⌝ = ⌜nf0⌝ ↔ σ0 = ⌜nf0⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%stream freely generated by ⌜ ⌝, ⌜ ⌝;;%len-02: ⊦ # a ' = 0 +1; from specification list%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%one-03: ⊦ a '[0] = a; from specification list-del%disj: ⊦ ⌜x⌝ ≠ ⌜nf⌝ ; from specification stream-basic