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Proof step 10 from proof append-to-gr-01 (previous step)

m0 = n0 + # x0 ∧ # x0 ≤ m0, # x0 < m0, ∀ m, σ. # x0 < m → ⌜x0⌝ + σ to m = ⌜x0⌝ + (σ to (m - # x0)) ∨ m = 0 ⊦ ⌜x0⌝ + σ to m0 = ⌜x0⌝ + (σ to n0), m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ m, σ. # x0 < m → ⌜x0⌝ + σ to m = ⌜x0⌝ + (σ to (m - # x0)) ∨ m = 0 ⊦ ⌜x0⌝ + σ to n0 + # x0 = ⌜x0⌝ + (σ to n0), n0 + # x0 = 0, n0 = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat