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Proof step 5 from proof append-to-le-01 (previous step)

∀ x0, a, σ. m0 ≤ # x0 + 1 ∧ (∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ to m = ⌜x0 to m⌝ ) → (⌜a '⌝ + ⌜x0⌝ ) + σ to m0 = ⌜a ' + x0 to m0⌝ , m0 +1 ≤ # x0 + 1, ∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ to m = ⌜x0 to m⌝ ⊦ (⌜a '⌝ + ⌜x0⌝ ) + σ to m0 +1 = ⌜a ' + x0 to m0 +1⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m0 ≤ # x0, ∀ x0, a, σ. m0 ≤ # x0 + 1 ∧ (∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ to m = ⌜x0 to m⌝ ) → (⌜a '⌝ + ⌜x0⌝ ) + σ to m0 = ⌜a ' + x0 to m0⌝ , ∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ to m = ⌜x0 to m⌝ ⊦ ⌜x0⌝ + σ to m0 = ⌜x0 to m0⌝

Rule argument

Additional information

Used simplifier rules: to-succ-cons: ⊦ ⌜a '⌝ + σ to n + 1 = a + (σ to n);%cons-append: ⊦ a + σ = ⌜a '⌝ + σ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%to-succ-cons-l-lem: ⊦ a ' + x to n + 1 = a + (x to n);%append-finfin: ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ;%append-cancel-l: ⊦ ⌜x⌝ + σ0 = ⌜x⌝ + σ1 ↔ σ0 = σ1;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2