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Proof step 87 from proof leninv-succ (previous step)

# nσ = ∞, m < m2, m < m1, m1 < m2, stutf(nσ), nσ[m] < nσ[m1], m + 1 < m2, m < m1 → nσ[m1] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ m1 = 0, nσ[m2] = nσ[m1], m2 = 0, nσ[m1] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 88:
nσ[m1] = nσ[m] + 1, # nσ = ∞, m + 1 < m2, nσ[m] < nσ[m1], stutf(nσ), m1 < m2, m < m1, m < m2, ∀ m0. m < m0 → nσ[m0] = nσ[m1] ⊦ nσ[m1] = 0, m2 = 0, nσ[m2] = nσ[m1], m1 = 0

Rule argument

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ler: ⊦ m ≤ m; from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat