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Proof step 19 from proof less-leninv (previous step)

# nσ = ∞, nσ[n] = nσ[m] + 1, nσ[n + 1] = nσ[n] + 1, m < n, nσ[m] < nσ[n], stutf(nσ), ∀ m1. m < m1 → nσ[m1] = nσ[n] ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n + 1 abort: F discard: T 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%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s) Used terms: n + 1