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Proof step 24 from proof inv-inverts (previous step)

# nσ = ∞, nσ[n] = nσ[m] + 1, nσ[n + 1] = nσ[n] + 1, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , m < n, stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[n] ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[n])[nσ[n]] = n

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n + 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s) Used terms: n + 1