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Proof step 50 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ < leninv(nσ), nσ[n1 +1] = nσ[n1] → ¬ n1 < n1 +1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = nσ[n1] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
# nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ < leninv(nσ), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = nσ[n1] + 1, nσ[n1 + 1] = nσ[n1]

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2