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Proof step 1 from proof short-sub-pos (conclusion)

⊦ s2i(sho) ≤ s2i(sho0) ∧ 0 ≤ s2i(sho) → s2i(i2s(s2i(sho0) - s2i(sho))) = s2i(sho0) - s2i(sho)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i - j = i0 ↔ i = i0 + j;