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Proof step 14 from proof del-three (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# s, m] flags:

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%size-one-02: ⊦ # s = 1 ∧ a ∈ s ∧ b ∈ s → a = b; (forward) from specification set-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # s, m