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Proof step 14 from proof and-ones (previous step)

∀ bits1, bits0. # bits1 < #(true + bits2) → true → bits1 = [] ∨ bitsand(bits1, false ' + bits0) = false ' + bits0 ∨ false ∈ bits1 ∨ false ∈ bits0 ⊦ bits0 = [], bits2 = [], false ∈ bits2, false ∈ bits0, (bits0.last → bits2.last) ∧ bitsand(true ' + bits2.butlast, false ' + bits0.butlast) = false ' + bits0.butlast

User applied rule apply induction

And obtained the following new goals

Proof step 15:
true ' + bits2.butlast = [] ∨ bitsand(true ' + bits2.butlast, false ' + bits0.butlast) = false ' + bits0.butlast ∨ false ∈ true ' + bits2.butlast ∨ false ∈ bits0.butlast, ∀ bits1, bits0. # bits1 < #(true + bits2) → true → bits1 = [] ∨ bitsand(bits1, false ' + bits0) = false ' + bits0 ∨ false ∈ bits1 ∨ false ∈ bits0 ⊦ bits0 = [], bits2 = [], false ∈ bits2, false ∈ bits0, (bits0.last → bits2.last) ∧ bitsand(true ' + bits2.butlast, false ' + bits0.butlast) = false ' + bits0.butlast

Rule argument

subst: [bits1, bits0] → [true ' + bits2.butlast, bits0.butlast]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ bits ≠ [] → # bits.butlast = # bits - 1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ m ≤ n ↔ n < m;%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: true ' + bits2.butlast, bits0.butlast