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Proof step 7 from proof Sep-forall (previous step)

∀ nP0, n, nP1. (∀ n0. n0 < m0 → nP0(n + n0) = nP1(n + n0)) → nP0 ⊗_(n, m0) = nP1 ⊗_(n, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) = nP1(n + n0) ⊦ nP0(n) * nP0 ⊗_(n + 1, m0) = nP1(n) * nP1 ⊗_(n + 1, m0)

User applied rule all left

And obtained the following new goals

Proof step 8:
(∀ n0. n0 < m0 → nP0((n + 1) + n0) = nP1((n + 1) + n0)) → nP0 ⊗_(n + 1, m0) = nP1 ⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) = nP1(n + n0) ⊦ nP0(n) * nP0 ⊗_(n + 1, m0) = nP1(n) * nP1 ⊗_(n + 1, m0)

Rule argument

pos: left 1 terms: nP0, n + 1, nP1 abort: F discard: T computed: F

Additional information

Used terms: nP0, n + 1, nP1