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Proof step 6 from proof append-from-le-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 ≤ # x0, ∀ x0, a, σ. m0 ≤ 1 + # x0 ∧ (∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ from m = ⌜x0 from m⌝ + σ) → (⌜a '⌝ + ⌜x0⌝ ) + σ from m0 = ⌜a ' + x0 from m0⌝ + σ, ∀ m, σ. m ≤ # x0 → ⌜x0⌝ + σ from m = ⌜x0 from m⌝ + σ ⊦ ⌜x0⌝ + σ from m0 = ⌜x0 from m0⌝ + σ

Rule argument

Additional information

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