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Proof step 8 from proof del-three (previous step)

# s = m + 1, m - 1 = 0, 2 < # s, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ # s = 0, b = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# s = m + 1, m - 1 = 0, a ∈ s, b ∈ s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b = a, m = 0, m = 1

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)