Natural Deduction
p & (q + r) <=> (p & q) + (p & r)
assume p & (q + r). p. q + r. assume q. p & q. p & q + p & r. therefore q => p & q + p & r. assume r. p & r. p & q + p & r. therefore r => p & q + p & r. p & q + p & r. therefore p & (q + r) => p & q + p & r. assume p & q + p & r. assume p & q. p. therefore p & q => p. assume p & r. p. therefore p & r => p. p. assume p & q. q. q + r. therefore p & q => q + r. assume p & r. r. q + r. therefore p & r => q + r. q + r. p & (q + r). therefore p & q + p & r => p & (q + r). p & (q + r) <=> p & q + p & r.
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Last Modified : 27-Feb-2023