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Proof step 3 from proof bytes2int-size (previous step)

2 ^ ((n→i(# bytes) + 1) * 8) ≤ bytes2int(i2b(0) ' + by ' + bytes) ∧ b2i(by) < 0, ∀ by, bytes0. # bytes0 < # bytes → 2 ^ ((n→i(# bytes0) + 1) * 8) ≤ bytes2int(i2b(0) ' + by ' + bytes0) ∧ b2i(by) < 0 → false ⊦ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
b2i(by) < 0, 2 ^ ((n→i(# bytes) + 1) * 8) ≤ bytes2int(i2b(0) ' + by ' + bytes), ∀ by, bytes0. # bytes0 < # bytes → 2 ^ ((n→i(# bytes0) + 1) * 8) ≤ bytes2int(i2b(0) ' + by ' + bytes0) ∧ b2i(by) < 0 → false ⊦

Proof step 3 from proof bytes2int-size (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information