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Proof step 9 from proof append-to-gr-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
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

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, # x0, n] flags:

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, # x0, n