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Proof step 8 from proof append-to-gr-01 (previous step)

m = m0 + 1, # x0 + 1 < m, ∀ m, σ. # x0 < m → ⌜x0⌝ + σ to m = ⌜x0⌝ + (σ to (m - # x0)) ∨ m = 0 ⊦ m = 0, ⌜x0⌝ + σ to m0 = ⌜x0⌝ + (σ to (m - # x0 + 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# 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

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2