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Proof step 6 from proof elim-fromto (previous step)

⌜x⌝ = ⌜x0⌝ + σ1, # x0 ≤ # x ⊦ x0 = x to # x0 ∧ σ1 = ⌜x from # x0⌝

User applied rule constructor cut

And closed the goal!

Rule argument

var: σ1

Additional information

Used simplifier rules: eq-stream-to-list-2-1: ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%append-to-size: ⊦ x + y to # x = x;%append-from-size: ⊦ x + y from # x = y;%neq-fin-inf-01: ⊦ ⌜x⌝ ≠ σ + ⌜nf0⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%stream freely generated by ⌜ ⌝, ⌜ ⌝;;%inj-02: ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; from specification stream-basic%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2