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Proof step 8 from proof del-three (
previous step
)
# s = m + 1, m - 1 = 0, a0 ∈ s, a ∈ s, b ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ m = 0, m = 1, b = a
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] → [m, m0] flags:
Additional information
Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0